Which Golf Cart Undergoes The Greater Magnitude Of Change In Momentum?
Momentum and Collisions Review
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Part E: Problem-Solving
57. A 0.530-kg basketball game hits a wall head-on with a forward speed of 18.0 m/southward. Information technology rebounds with a speed of thirteen.5 m/due south. The contact time is 0.100 seconds. (a) make up one's mind the impulse with the wall, (b) decide the strength of the wall on the brawl.
Reply: Answer: (a) -xvi.seven N s; (b) -167 N
Given: grand = 0.530 kg; 5i = 18.0 m/south; 5f = 13.5 yard/s; t = 0.100 south
Notice: (a) Impulse, (b) Strength
(a) Impulse = Momentum Alter = 1000•Delta v = 1000•(vf - vi)= (0.530 kg)•( -thirteen.5 yard/s - 18.0 k/south)
Impulse = -16.7 kg•m/s = -xvi.7 N•southward
where the "-" indicates that the impulse was opposite the original direction of movement.
(Note that a kg•thousand/due south is equivalent to a N•south)
(b) The impulse is the product of strength and time. And so if impulse is known and time is known, strength can be hands determined.
Impulse = F•t
F = Impulse/t = (-16.7 North s) / (0.100 south) = -167 Northward
where the "-" indicates that the impulse was opposite the original direction of motility.
58. A 4.0-kg object has a forward momentum of twenty. kg•chiliad/s. A 60. Due north•s impulse acts upon it in the direction of motion for v.0 seconds. A resistive strength of 6.0 N then impedes its motion for viii.0 seconds. Make up one's mind the final velocity of the object.
Answer: 5f = 8.0 m/southward
This question is best idea most conceptually using the principle that an objects momentum is changed when information technology encounters an impulse and the corporeality of change in momentum is equal to the impulse which it encounters.
Here an object starts with 20 units (kg•thou/s) of momentum. It and then encounters an impulse of 60 units (Due north•due south) in the management of motion. A lx-unit impulse will change the momentum by 60 units, either increasing or decreasing it. If the impulse is in the direction of an object'due south motion, then it will increment the momentum. And so now the object has 80 units (kg•m/s) of momentum. The object then encounters a resistive force of 6.0 Due north for viii.0 s. This is equivalent to an impulse of 48 units (North•southward). Since this impulse is "resistive" in nature, it will decrease the object's momentum by 48 units. The object now has 32 units of momentum. The question asks for the object's velocity after encountering these two impulses. Since momentum is the product of mass and velocity, the velocity can be easily determined.
vfinal = pfinal / m = (32 kg chiliad/s) / (4.0 kg) = 8.0 m/southward
59. A 3.0-kg object is moving forrad with a speed of vi.0 chiliad/s. The object then encounters a force of 2.5 N for 8.0 seconds in the direction of its move. The object and then collides caput-on with a wall and heads in the opposite management with a speed of 5.0 m/s. Determine the impulse delivered past the wall to the object.
Reply: 53 N•s
Like the previous trouble, this problem is best solved past thinking through it conceptually using the impulse-momentum change principle.
Here the object begins with a momentum of 18 units (kg•thou/s). The object encounters a force of 2.5 N for 8.0 seconds. This is equivalent to an impulse of 20 units (N•s). Since this impulse acts in the direction of motility, it changes the object'south momentum from eighteen units to 38 units. A final impulse is encountered when colliding with a wall. Upon rebounding, the object has a momentum of -15 units (kg•m/s). The -15 is the product of mass (3 kg) and velocity (-v thou/due south). The "-" sign is used since the object is now moving in the reverse direction as the original motion. The standoff with the wall changed the object's momentum from +38 units to -xv units. Thus, the standoff must have resulted in a 53-unit impulse since information technology altered the object's momentum by 53 units.
60. A 46-gram lawn tennis ball is launched from a 1.35-kg homemade cannon. If the cannon recoils with a speed of 2.i m/s, make up one's mind the muzzle speed of the tennis ball.
Answer: 62 one thousand/s
Given: yardball = 46 one thousand = 0.046 kg; mcannon = i.35 kg; 5cannon = -2.1 m/southward
Find: vbrawl = ???
The ball is in the cannon and both objects are initially at rest. The full organisation momentum is initially 0. After the explosion, the full system momentum must also be 0. Thus, the cannon's backward momentum must be equal to the brawl'south forward momentum.
(1.35 kg) • (-two.1 one thousand/south) = (0.046 kg) • vball
vbrawl = (1.35 kg) • (two.i m/due south) / (0.046 kg) = 61.63 thousand/s = ~62 m/s
61. A 2.0-kg box is attached by a string to a 5.0-kg box. A compressed spring is placed between them. The two boxes are initially at rest on a friction-gratis track. The string is cutting and the spring applies an impulse to both boxes, setting them in move. The 2.0-kg box is propelled backwards and moves 1.2 meters to the finish of the track in 0.l seconds. Determine the time it takes the 5.0-kg box to move 0.90 meters to the reverse end of the track.
Answer: 0.94 s
For the sake of the word, the 2-kg box will exist referred to equally Box 1 and the five-kg box will be referred to as box ii.
Given: mbox one = two.0 kg; yardbox two = 5.0 kg; dbox 1 = 1.2 thousand; tbox 1 = 0.50 south; dbox ii = 0.xc g
Find: tbox 2 = ???
The ii boxes are initially at rest. The total system momentum is initially 0. After the cutting of the string and the impulse of the leap, the total arrangement momentum must also be 0. Thus, Box i's backward momentum must exist equal to the Box 2's forward momentum. The distance and time for Box one must be used to determine its velocity.
Now the principle of momentum conservation can be used to determine Box ii'south velocity.
(2 kg) • (ii.four k/s) = (5 kg) • 5box 2
vbox two = (ii kg) • (2.four m/due south) / (five kg) = 0.96 m/southward
The velocity of Box 2 tin can exist used to make up one's mind the time it takes it to move a altitude of 0.ninety meters.
Fourth dimension = dbox 2 / vbox 2 = (0.90 chiliad) / (0.96 m/s) = 0.9375 s = ~0.94 s
62. Two children are playing with a large snowball while on water ice skates on a frozen pond. The 33-kg child tosses the five.0-kg snowball, imparting a horizontal speed of 5.0 chiliad/s to information technology. The 33-kg child is 4.0 meters from a 28-kg kid and eight.0 meters from the edge of the pond (located behind him). Assuming negligible friction, how much time elapses betwixt when the 28-kg child gets hit past the snowball and when the 33-kg child reaches the edge of the pond?
Answer: 9.8 s
For the sake of the discussion, nosotros will refer to the 33-kg child as the "thrower" and the 28-kg kid every bit the "catcher."
In this scenario, the thrower tosses a snowball forward towards the catcher. This throwing action involves an impulse imparted to the snowball. And due to action-reaction, there is an identical impulse imparted to the thrower which causes the thrower to be set in movement in the contrary direction. The impulse is equal to the momentum change. And since the mass and the velocity change of the snowball are known, the momentum change of the snowball tin be determined.
yard • (Delta 5)snowball = (five.0 kg) • (+ 5.0 m/southward - 0 grand/s) = 25.0 kg•m/s
This 25-unit momentum change of the snowball is equal to the thrower's momentum alter. Thus the thrower is moving backwards towards the edge of the pond with a momentum of -25.0 kg•m/s. Since momentum is related to velocity, the post-impulse velocity can exist determined.
vthrower = pthrower / mthrower
5thrower = (-25.0 kg•chiliad/s) / (33 kg)
vthrower = -0.7576 thousand/s
The thrower began 8.0 meters from the edge of the pond. Once the ball has been thrown, the thrower is moving backwards towards the border of the pond with a speed of 0.7576 thou/s. Assuming negligible friction on the icy swimming, the speed tin can exist used to determine the fourth dimension that elapses between when the brawl is thrown and when the thrower reaches the swimming'south edge.
t = d / 5
tfor thrower to reach pond's border = (viii.0 meters) / (0.7576 m/due south)
tfor thrower to reach pond's border = 10.56 seconds
Once the ball is thrown, the thrower starts moving backwards towards the pond'due south edge. Meanwhile, the ball is moving forward towards the catcher. The time for the ball to move from the thrower's original position to the catcher is dependent upon the ball'due south speed and the original altitude between the thrower and the catcher.
t = d / v
tfor ball to move from thrower to catcher = (iv.0 meters) / (5.0 thousand/s)
tfor ball to motion from thrower to catcher = 0.800 seconds
The instant the ball is thrown, ii motions occur - the ball moves forward towards the catcher and the thrower moves backwards towards the pond's edge. The ball reaches the catcher in 0.800 seconds. And 9.76 seconds (~nine.8 s) later (ten.56 s - 0.fourscore s), the thrower reaches the pond'due south edge.
63. A two.8-kg physics cart is moving forward with a speed of 45 cm/s. A 1.9-kg brick is dropped from residue and lands on the cart. The cart and brick move together beyond the horizontal surface. Assume an isolated system.
a. Determine the mail service-collision speed of the cart and the brick.b. Determine the momentum alter of the cart.
c. Decide the momentum alter of the brick.
d. Make up one's mind the cyberspace impulse upon the cart.
e. Determine the cyberspace impulse upon the system of cart and brick.
Answers: (a) 5 = 27 cm/s
(b) Delta pcart = -51 kg • cm/s
(c) Delta pbrick = +51 kg • cm/southward
(d) Impulse on cart = -51 kg • cm/south
(eastward) Impulse on brick = +51 kg • cm/s
Before the collision, simply the moving cart has momentum. The total momentum of the organization is but the mass of the cart multiplied by the velocity of the cart.
ptotal-before = (2.viii kg) • (45 cm/s)
ptotal-earlier = 126 kg • cm/s
The collision is perfectly inelastic; the two objects stick together and motion equally a single unit of measurement. After the collision, the total momentum of the organisation is the sum of the individual momentum values.
pfull-later on = mcart-after • vcart-afterward + mbrick • fivebrick-subsequently
ptotal-after = mcart-after • v + mbrick • v
ptotal-afterwards = (ii.8 kg) • v + (ane.ix kg) • five
ptotal-after = (4.7 kg) • v
Assuming an isolated arrangement, total system momentum is conserved. Thus, before- and later-collision momentum expressions can be set equal to each other, and the equation can be manipulated to solve for the post-standoff speed of the two objects.
v = (126 kg • cm/due south) / (4.7 kg)
five = 26.809 cm/due south (~27 cm/s)
The momentum change of the cart (Delta pcart) is simply the difference betwixt the initial and final momentum values.
Delta pcart = mcart-after • fivecart-afterwards - mcart • vcart-earlier
Delta pcart = (2.8 kg) • (26.809 m/southward) - (two.8 kg) • (45 cm/s)
Delta pcart = -50.936 kg • cm/s (~51 kg•cm/s)
During the collision, the cart loses 50.nine units of momentum. Since total organisation momentum is conserved, the brick must gain the same quantity of momentum.
The modify in momentum of the cart is due to the fact that an impulse acts upon the cart during the collision. The moment contact is made between the brick and the cart, the 2 objects are moving at unlike speeds relative to each other. Consequently, there will be a friction strength acting between the two surfaces until the two objects maintain the same speed. That is, the brick will pull backward upon the moving cart in order to dull it down; and the moving cart will pull forward upon the stationary brick in order to speed information technology upwards. These impulses are what crusade the momentum changes. And these impulses are equal to the momentum changes. Thus,
Impulse on brick = Delta pbrick = +50.936 kg • cm/s (~51 kg•cm/s)
64. In a physics lab, a 0.500-kg cart moving at 36.4 cm/s collides inelastically with a 2d cart which is initially at rest. The ii carts move together with a speed of 21.8 cm/s after the collision. Determine the mass of the second cart.
Answer: ~ 0.335 kg
This trouble involves a perfectly inelastic collision between two carts. Thus, the post-collision velocity of the 2 carts are identical. For advice sake, the carts will exist referred to equally Cart A and Cart B. The given information is:
The unknown to be solved for in this problem is the mass of Cart B (gB).
The solution begins by setting the writing expressions for the total momentum of the organisation before and after the collision.
Before Collision: pfull-before = (0.500 kg)•(36.four cm/s) + (mB)•(0 cm/south)Afterwards Standoff: ptotal-subsequently = (0.500 kg)•(21.viii cm/southward) + (grandB)•(21.8 cm/southward)
Assuming momentum conservation, these expressions are gear up equal to each other so algebraically manipulated to solve for the unknown (mB).
(0.500 kg)•(36.4 cm/s) = (0.500 kg)•(21.viii cm/s) + (gB)•(21.8 cm/south)
(0.500 kg)•(36.4 cm/south) - (0.500 kg)•(21.8 cm/s) = (grandB)•(21.viii cm/southward)
(7.xxx kg•cm/s) = (mB)•(21.8 cm/s)
0.33486 kg = mB
mB = ~0.335 kg
65. A 9230-kg truck collides head on with a 1250-kg parked machine. The vehicles entangle together and slide a linear distance of 10.half-dozen meters earlier coming to rest. Assuming a uniform coefficient of friction of 0.820 between the road surface and the vehicles, determine the pre-collision speed of the truck.
Answer: 14.8 m/s
Here is an case of a more than difficult trouble involving the combination of momentum principles with information learned in other units of the class. The trouble involves a perfectly inelastic standoff betwixt a truck and a automobile. Thus, the post-collision velocity of the truck and car are identical. The given collision data is:
The unknown to be solved for in this problem is the velocity of the truck before the collision (vTruck-before).
There is some kinematic/dynamic information provided that will help in determining the post-collision velocity of the entangled truck and car.
The combination of automobile and truck will slide to a final resting position due to the action of friction. The coefficient of friction, a free-body diagram and a kinematic equation tin can be used to determine the velocity of the auto and truck immediately following the collision.
The free-body diagram is shown at the correct. Note that the unbalanced force is friction. Its value is found past multiplying the coefficient of friction by the combined weight of the car and truck (mu•Chiliad•g where M = grandtruck + mautomobile ). The forcefulness of friction is the cyberspace force. Thus, the acceleration is the force of friction divided past the combined mass of the motorcar and the truck (M). Subsequently, the expression for the acceleration of the car and truck while sliding to a stop is but mu•g.
a = -8.036 m/s/south
(The - sign indicates a deceleration or slowing down motion.)
Now a kinematic equation can be used to solve for the velocity of the car and truck immediately after the collision. This is shown below:
(0 m/s)2 = vo 2 + 2•(-8.036 1000/southward/s)•(ten.6 yard) = fiveo ii - 170.36 yard2/s2
(0 m/s)2 = vo ii - 170.36 m2/south2
vo 2 = 170.36 mii/s2
vo = SQRT(170.36 m2/s2)
vo = 13.052 g/s (= fiveTruck-after = vAuto-after )
Now that the post-standoff velocity of the automobile and truck are known, expressions for the full system momentum can be written for the before- and subsequently-collision situations.
Before Standoff: pfull-before = (9320 kg)•(vTruck-before) + (1250 kg)•(0 cm/due south)After Collision: ptotal-after = (9320 kg)•(thirteen.052 m/south) + (1250 kg)•(xiii.052 m/southward)
Assuming momentum conservation, these expressions are prepare equal to each other and then algebraically manipulated to solve for the unknown (mB).
(9320 kg)•(vTruck-before) = (9320 kg)•(13.052 yard/south) + (1250 kg)•(xiii.052 g/due south)
(9320 kg)•(fiveTruck-before) = 137963 kg•m/s
(vTruck-before) = (137963 kg•m/s) / (9320 kg)
vTruck-earlier = ~14.8 yard/s
66. A archetype physics sit-in involves firing a bullet into a block of forest suspended by strings from the ceiling. The peak to which the woods rises below its everyman position is mathematically related to the pre-collision speed of the bullet. If a 9.7-gram bullet is fired into the center of a i.1-kg cake of wood and it rises up a distance of 33 cm, then what was the pre-collision speed of the bullet?
Reply: 2.nine 10 102 one thousand/s
Here is another case in which momentum principles must be combined with content learned in other units in order to complete an analysis of a concrete situation. The collision involves the inelastic collision betwixt a block of wood and bullet. The bulled lodges into the wood and the ii objects move with identical velocity after the collision. The kinetic free energy of the forest and bullet is then converted to potential energy as the combination of two objects rises to a concluding resting position.
Energy conservation can be used to make up one's mind the velocity of the forest-bullet combination immediately after the collision. The kinetic energy of the wood-bullet combination is set equal to the final potential free energy of the wood-bullet combination and the equation is manipulated to solve for the post-collision velocity of the wood-bullet combination. The work is shown here:
0.5 • vcombination-after 2 = (9.8 m/sii) • (0.33 grand)
vcombination-afterward 2 = 2 •(ix.8 m/due south2) • (0.33 m)
vcombination-subsequently 2 = 6.468 m2/s2
vcombination-subsequently = two.5432 thousand/s
Now momentum conservation can be used to determine the pre-collision velocity of the bullet (vbullet-earlier). The known information is:
Expressions for the total organization momentum can exist written for the earlier- and after-collision situations.
Before Standoff: ptotal-earlier = (1.1 kg)•(0 m/s) + (0.0097 kg)•(vbullet-before)Later Collision: ptotal-later = (1.1 kg)•(ii.5432 g/s) + (0.0097 kg)•(2.5432 1000/southward)
Assuming momentum conservation, these expressions are gear up equal to each other and and then algebraically manipulated to solve for the unknown (grandB).
(0.0097 kg)•(vbullet-before) = (1.1 kg)•(2.5432 1000/s) + (0.0097 kg)•(2.5432 thou/s)
(0.0097 kg)•(vbullet-earlier) = 2.8222 kg•m/s
fivebullet-earlier = (2.8222 kg•m/s) / (0.0097 kg)
vbullet-before = 290.95 chiliad/s
fivebullet-before = ~ii.9 10 10two grand/south
67. At an entertainment park, twin brothers Timmy (thou = l kg) and Jimmy (grand = 62 kg) occupy separate 36-kg bumper cars. Timmy gets his car cruising at 3.6 m/s and collides head-on with Jimmy who is moving the opposite direction at 1.6 k/due south. Subsequently the collision, Timmy bounces backwards at 0.five chiliad/southward. Assuming an isolated system, determine ...
a. ... Jimmy's post-collision speed.b. ... the percentage of original kinetic energy which is lost as the issue of the collision.
Answer: (a) v = ~2.0 thou/s
(b) % KE Loss = ~lxx. %
(a) Expressions for the total momentum of the system before and after the collision can be written. For the earlier-collision expression, Timmy is assigned a positive velocity value and Jimmy is assigned a negative velocity value (since he is moving in the reverse direction). Furthermore, the mass of the bumper automobile must be figured into the total mass of the individually moving objects.
ptotal-earlier = mTimmy • vTimmy-before + mJimmy • 5Jimmy-before
ptotal-before = (86 kg) • (3.6 yard/due south) + (98 kg) • (-1.half dozen m/due south)
For the earlier-collision expression, Timmy is assigned a negative velocity value (since he has bounced backwards in the opposite direction of his original motion. Jimmy is assigned a velocity of v since his velocity is not known.
ptotal-after = kTimmy • vTimmy-afterward + thousandJimmy • vJimmy-later on
ptotal-later on = (86 kg) • (-0.5 thousand/s) + (98 kg) • (vJimmy-after)
pfull-after = (86 kg) • (-0.five 1000/south) + (98 kg) • v
Since the system is assumed to be isolated, the before-collision momentum expression tin exist ready equal to the after-standoff momentum expression. The equation tin then be algebraically manipulated to solve for the post-standoff velocity of Jimmy.
309.6 kg•m/south - 156.8 kg•m/south = (86 kg) • (-0.5 one thousand/s) + (98 kg) • v
152.8 kg•m/due south = - 43 kg•grand/s + (98 kg) • v
152.eight kg•m/s + 43 kg•thou/south = (98 kg) • v
195.viii kg•thou/due south = (98 kg) • v
5 = (195.8 kg•m/due south) / (98 kg)
v = one.998 grand/south = ~2.0 g/south
(b) This collision is neither perfectly elastic (since the collision forcefulness is a contact force) nor perfectly inelastic (since the objects do not stick together). It is a partially elastic/inelastic standoff. Since the collision is not perfectly elastic, there is a loss of full organisation kinetic energy during the standoff. The before-collision and afterward-collision kinetic energy values can be calculated and the percentage of total KE lost can be determined.
The before-standoff KE is based on before-collision speeds:
KEsystem-before = 0.5 • gTimmy • vTimmy -before 2 + 0.5 • kJimmy • vJimmy-before 2
KEarrangement-before = 0.5 • (86 kg) • (3.6 m/s)2 + 0.v • (98 kg) • (one.6 m/s)ii
KEsystem-before = 557.28 J + 125.44 J
KEorganization-before = 682.72 J
The after-collision KE is based on afterward -collision speeds:
KEsystem-subsequently = 0.five • grandTimmy • vTimmy-later + 0.5 • grandJimmy • fiveJimmy-afterward
KEsystem-after = 0.5 • (86 kg) • (0.5 m/s)two + 0.5 • (98 kg) • (1.998 k/s)2
KEsystem-subsequently = x.75 J + 195.61 J
KEarrangement-after = 206.36 J
The arrangement kinetic free energy is changed from 682.72 J to 206.36 J during the standoff. The total KE lost is 476.36 J. This value can be used to decide the percentage of the original KE which is lost in the collision.
% KE Loss = 69.77% = ~70%
68. Two billiard balls, assumed to have identical mass, collide in a perfectly elastic collision. Ball A is heading E at 12 m/s. Ball B is moving West at viii.0 g/south. Make up one's mind the post-collision velocities of Ball A and Brawl B.
Respond: vA-after = -viii.0 cm/s; vB-after = 12 cm/s (The - indicates West and the + indicates East)
This standoff is said to be perfectly rubberband. Thus, both the total system momentum and the total system kinetic energy of the 2 objects is conserved. The momentum conservation equation tin be written as
Since the balls are identical, their masses are the same. That is, gA = thousandB = yard. The equation can be rewritten as:
Since each term of the equation contains the variable thou, we tin can divide through by m and cancel m'due south from the equation. The equation can be rewritten as:
vA-before + vB-earlier = vA-after + vB-after For elastic collisions, full system kinetic free energy is conserved. The kinetic energy conservation equation is written as
As shown in the book, this equation can exist simplified to the form of
vA-before + vA-after = 5B-earlier + 5B-after The problem states the earlier-collision velocities of the two balls.
vB-before = - eight cm/due south (the - indicates west)
These two values can be substituted into equations 1 and 2 above.
12 cm/southward - eight cm/southward = vA-after + fiveB-after 
12 cm/s + vA-subsequently = - 8 cm/due south + vB-after
Now the problem has been reduced to two equations and two unknowns. Such a problem can be solved in numerous ways. One method involved using Equation three to develop an expression for 5A-subsequently in terms of vB-after. This expression for fiveA-afterwards can then be substituted into Equation 4. The value of vB-afterward can and then exist determined. This is shown below.
From Equation 3:
vA-afterward = iv cm/s - 5B-later
This expression for fiveA-subsequently in terms of fiveB-after can at present be substituted into equation 4. This is shown beneath. The subsequent algebraic manipulation is shown also.
12 cm/s + 4 cm/southward + 8 cm/s - vB-after = + vB-later on
24 cm/south = + vB-after + 5B-subsequently
24 cm/s = two vB-later
fiveB-after = +12 cm/s
Now that the value of vB-afterwards has been determined, it tin can be substituted into the original expression for vA-afterwards (Equation 5) in order to make up one's mind the numerical value of fiveA-after. This is shown beneath.
vA-later = four cm/southward - 12 cm/s
vA-after = -8.0 cm/due south
69. A 1.72-kg cake of soft woods is suspended by 2 strings from the ceiling. The wood is costless to rotate in pendulum-like fashion when a force is exerted upon information technology. A viii.fifty-g bullet is fired into the forest. The bullet enters the wood at 431 m/s and exits the contrary side shortly thereafter. If the wood rises to a meridian of thirteen.eight cm, and then what is the go out speed of the bullet?
Respond: vbullet-later = 98.2 m/s
The difficulty of this trouble lies in the fact that information from other units (work and free energy) must be combined with the momentum data from this unit to arrive at a solution to the problem. In this scenario there is a collision between a stationary cake of woods and a moving bullet. The impulse causes the block of forest to be set into motion and the bullet to tiresome down. Momentum tin can be causeless to be conserved. Once gear up into motion, the block of wood rises in pendulum-similar style to a given pinnacle. Its free energy of motion (kinetic energy) is transformed into energy of vertical position (potential energy). The mail service-standoff speed of the wood can be determined using energy conservation equations.
To begin the solution, the final height of the forest is used to determine the mail-standoff speed of the forest.
0.5 • mforest •vwood 2 = one thousandwood • m • hwood
fivewood 2 = 2 • g • hwood
fiveforest = SQRT(ii • thou • hwood)
vforest = SQRT[2 • (9.viii m/southtwo) • (0.138 k)]
vwood = SQRT[2.7048 mii/s2]
vwood = 1.6446 m/s
Immediately following the emergence of the bullet from the wood, the wood cake is moving with a speed of ane.6446 grand/s. Knowing this, momentum conservation tin can be applied to make up one's mind the post-standoff speed of the bullet.
where fivewood-before = 0 m/s; 5bullet-before = 431 m/s; 5wood-later = i.6446 k/s; vbullet-after = ???
(1.72 kg) • (0 chiliad/s) + (0.00850 kg) • (431 m/s) = (one.72 kg) • (ane.6446 m/s) + (0.00850 kg) • vbullet-after
(To simplify the work, the units will exist dropped from the solution in the next several steps. Once a vbullet-after value is found, its units will exist in thousand/due south, consequent with the units stated in the above line.)
0 + 3.6635 = ii.8288 + 0.00850 • vbullet-after
0.8347 = 0.00850 • vbullet-after
(0.8347) / (0.00850) = fivebullet-after
vbullet-afterward = 98.205 m/south = ~98.ii chiliad/south
70. In a physics lab, the pitching speed of a student is determined past throwing a baseball into a box and observing the box'south motion later on the catch. A measurement of the the altitude the box slides beyond a rough surface of known coefficient of friction volition allow i to make up one's mind the pre-impact speed of the pitched brawl. If a 0.256-kg ball hits a iii.46-kg box and the ball and box slide a distance of 2.89 meters beyond a surface with a coefficient of friction of 0.419, then what is the pre-impact speed of the pitched ball?
Answer: 70.7 m/s
This is another example of a problem in which information from other units (work and free energy or Newton's laws and kinematics) must be combined with the momentum information from this unit to arrive at a solution to the problem. At that place is a collision between a stationary box and a moving baseball that causes the baseball game to slow down and the box to speed upwards. It is a perfectly inelastic collision with the baseball game remaining lodged in the box and the two objects moving together with the same post-collision speed. After the collision occurs, the baseball and box slide a given altitude across a rough surface to a terminal resting position. The coefficient of friction between the box and the surface is given. This latter information (sliding altitude and mu value) can be used to determine the post-collision speed of the box and baseball. Once found, momentum conservation can be applied to the collision to make up one's mind the pre-collision speed of the baseball.
Work and energy principles volition exist used to analyze the motion of the box/baseball organisation sliding to a stop. (Newton's laws and kinematics could just as easily been used). Immediately following the collision, the box/baseball system has kinetic energy. Friction does work upon the box/baseball arrangement to bring to a final resting position, characterized by zero kinetic energy. The motion occurs across a level surface, so there is no potential energy change of the box. The work done by friction is equal to the kinetic energy alter of the box/baseball system.
Ffrict • d • cos(Theta) = KEfinal - KEinitial
(mu •Fnorm) • d • cos(180) = 0.5 • m •vfinal 2 - 0.v • m •fiveinitial 2
- (mu • k • g) • d = 0 - 0.5 • m • 5initial 2
mu • yard • d = 0.five • fiveinitial 2
2 •mu • g • d = vinitial 2
SQRT(2 • mu • g • d) = vinitial
where mu = 0.419; d = two.89 m; and g = 9.8 m/stwo
fiveinitial = SQRT[2 • (0.419) • (ix.8 one thousand/s2 ) • (2.89 g)]
vinitial = SQRT[23.734 grand2/sii]
vinitial = 4.8717 1000/due south
Immediately following the collision, the box/baseball arrangement begins moving with a speed of 4.8717 m/s. Now momentum conservation can be applied to make up one's mind the pre-standoff speed of the baseball.
(0.256 kg) • vbaseball-before + (iii.46 kg) • (0 m/s) = (0.256 kg) • (4.8717 m/s) + (3.46 kg) • (4.8717 k/due south)
(To simplify the solution, the units will be dropped from the solution in the side by side several steps. Once a 5baseball game-before value is found, its units will be in yard/s, consistent with the units stated in the above line.)
0.256 • vbaseball game-earlier = 1.2472 + 16.8562
0.256 • vbaseball-earlier = xviii.1033
vbaseball-before = (eighteen.1033) / (0.256)
vbaseball-before = 70.seven m/s
71. Two ice skaters collide on the ice. A 39.vi-kg skater moving South at 6.21 thou/s collides with a 52.one-kg skater moving East at iv.33 one thousand/s. The two skaters entangle and move together across the water ice. Determine the magnitude and management of their mail-collision velocity.
Answer: 3.64 m/s at 42.five degrees east of due south (312.5 degrees)
The difficulty of this trouble lies in the fact that the standoff occurs between two objects moving at right angles to each other. Thus, vector principles will have to be combined with momentum principles to make it at a solution to the problem. The same conservation of momentum principle will be used; only when summing the before momentum values of the ii objects, the fact that they are at right angles to each means that they will have to exist added using the Pythagorean theorem. The standoff is perfectly inelastic with the 2 skaters moving at the same speed after the collision. For communication sake, the 39.half dozen-kg skater will be referred to as skater A and the 52.ane-kg skater will exist referred to as skater B. A vector diagram will probable assist in the solution of the trouble.
The individual momentum of the ii skaters is kickoff determined.
pB = mB • B = (52.1 kg) • (4.33 m/due south, East ) = 225.593 kg • k/southward, East
At present the Pythagorean theorem can be used to add these ii vectors and thus make up one's mind the pre-collision system momentum. The diagram at the right shows the vectors being added in head-to-tail way. The resultant is fatigued from the tail of the get-go vector to the head of the concluding vector. The resultant is the hypotenuse of a right triangle whose sides are the pA and pB vectors.
parrangement = pA + pB (where pA and pB are correct bending vectors) psystem = SQRT(pA 2+ pB 2)
porganisation = SQRT[(245.916 kg • m/s)2+ (225.593 kg • m/south)ii]
psystem = 333.717 kg • m/s
The management of the this total system momentum vector can be determined by using a trigonometric role. Equally shown in the diagram above, the angle theta is the bending between the system momentum vector and the vertical. This bending tin can be adamant using either the tangent, cosine or sine function. The tangent function is used below.
tangent(Theta) = pB / pA
tangent(Theta) = (225.593 kg • yard/s) / (245.916 kg • m/s)
tangent(Theta) = 0.91735
Theta = tan-1 (0.91735)
Theta = 42.532 degrees
Before the standoff, the total system momentum is 333.717 kg • m/s in a direction of 42.532 degrees east of south. Since total system momentum is conserved, the subsequently-collision momentum of the organisation is likewise 333.717 kg • yard/due south in a direction of 42.532 degrees east of s. After the standoff, the two objects move together as a unmarried unit of measurement with the same velocity. The velocity of each object can be found past dividing the total momentum by the total mass.
5system = ( porganization ) / (msystem )
5system = (333.717 kg • thou/s) / (91.vii kg)
5system = 3.64 m/s at 42.5 degrees e of southward
72. In a physics lab, two carts collide elastically on a level, low-friction track. Cart A has a mass of 1.500 kg and is moving due east at 36.5 cm/south. Cart B has a mass of 0.500 kg and is moving West at 42.8 cm/south. Determine the mail-collision velocities of the two carts.
Respond: fiveA-after = -3.xv cm/due south; fiveB-after = 76.xv cm/south
This is a perfectly elastic collision in which both momentum and kinetic energy are conserved. The method for solving this problem will be very similar to that used in Problem #68 above. Two equations will be developed using the momentum conservation and kinetic energy conservation principles. I equation volition be used to develop an expression for 5A in terms of 5B. This expression volition then be substituted into the second equation in order to solve for vB. The original 5A expression can and then be used to decide the vA value. The solution is shown below.
The momentum conservation equation can be written as
(ane.500 kg) • (+36.five cm/southward) + (0.500 kg) • (-42.eight cm/southward) = (1.500 kg) • vA-subsequently + (0.500 kg) • vB-subsequently
33.35 kg • cm/s = (1.500 kg) • vA-afterwards + (0.500 kg) • 5B-after
For elastic collisions, total arrangement kinetic energy is conserved. The kinetic energy conservation equation is written equally
As shown in the volume, this equation can be simplified to the grade of
36.5 cm/southward + vA-after = -42.8 cm/s + vB-later
5A-afterwards = 5B-later on - 79.three cm/s
At present the problem has been reduced to two equations and two unknowns. Such a problem tin can be solved in numerous ways. Annotation that equation two represents an expression for vA-after in terms of vB-afterwards. This expression for 5A-after tin then exist substituted into Equation 1. The value of vB-afterward can then be determined. This work is shown beneath. (To simplify the mathematics, the units will be dropped from the numerical values stated in the solution. When vB-after is solved for, its units will be in cm/s - the same units used for velocity in the above portion of the solution.)
33.35 = i.500 • vB-later - 118.95 + 0.500 • vB-later on
152.30 = 2.00 • vB-after
vB-afterward = 76.15 cm/s
Now that the value of vB-after has been determined, information technology can exist substituted into the original expression for vA-later on (Equation ii) in order to determine the numerical value of vA-after. This is shown beneath.
vA-subsequently = 76.15 cm/s - 79.three cm/s
vA-after = -3.15 cm/s
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