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Calculating final velocity and energy release: two carts collide

In the collision pictured in [link] , two carts collide inelastically. Cart 1 (denoted m 1 size 12{m rSub { size 8{1} } } {} carries a spring which is initially compressed. During the collision, the spring releases its potential energy and converts it to internal kinetic energy. The mass of cart 1 and the spring is 0.350 kg, and the cart and the spring together have an initial velocity of 2 . 00 m/s size 12{2 "." "00"`"m/s"} {} . Cart 2 (denoted m 2 size 12{m rSub { size 8{2} } } {} in [link] ) has a mass of 0.500 kg and an initial velocity of 0 . 500 m/s size 12{ - 0 "." "500"`"m/s"} {} . After the collision, cart 1 is observed to recoil with a velocity of 4 . 00 m/s size 12{ - 4 "." "00"`"m/s"} {} . (a) What is the final velocity of cart 2? (b) How much energy was released by the spring (assuming all of it was converted into internal kinetic energy)?

Strategy

We can use conservation of momentum to find the final velocity of cart 2, because F net = 0 size 12{F rSub { size 8{"net"} } =0} {} (the track is frictionless and the force of the spring is internal). Once this velocity is determined, we can compare the internal kinetic energy before and after the collision to see how much energy was released by the spring.

Solution for (a)

As before, the equation for conservation of momentum in a two-object system is

m 1 v 1 + m 2 v 2 = m 1 v 1 + m 2 v 2 .

The only unknown in this equation is v 2 . Solving for v 2 and substituting known values into the previous equation yields

v 2 = m 1 v 1 + m 2 v 2 m 1 v 1 m 2 = 0.350 kg 2.00 m/s + 0.500 kg 0.500 m/s 0.500 kg 0.350 kg 4.00 m/s 0.500 kg = 3.70 m/s.

Solution for (b)

The internal kinetic energy before the collision is

KE int = 1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 = 1 2 0 . 350 kg 2.00 m/s 2 + 1 2 0 . 500 kg 0 . 500 m/s 2 = 0 . 763 J .

After the collision, the internal kinetic energy is

KE int = 1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 = 1 2 0.350 kg - 4.00 m/s 2 + 1 2 0.500 kg 3.70 m/s 2 = 6.22 J.

The change in internal kinetic energy is thus

KE int KE int = 6.22 J 0 . 763 J = 5.46 J.

Discussion

The final velocity of cart 2 is large and positive, meaning that it is moving to the right after the collision. The internal kinetic energy in this collision increases by 5.46 J. That energy was released by the spring.

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Test prep for ap courses

Mass A (2.0 kg) has an initial velocity of 4 m/s in the + x -direction. Mass B (2.0 kg) has an initial velocity of 5 m/s in the – x -direction. If the two masses have an elastic collision, what will be the final velocities of the masses after the collision?

  1. Both will move 0.5 m/s in the – x -direction.
  2. Mass A will stop; mass B will move 9 m/s in the + x -direction.
  3. Mass B will stop; mass A will move 9 m/s in the – x -direction.
  4. Mass A will move 5 m/s in the – x -direction; mass B will move 4 m/s in the + x -direction.

(d)

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Mass A has an initial velocity of 22 m/s in the + x -direction. Mass B is three times more massive than mass A and has an initial velocity of 22 m/s in the – x -direction. If the two masses have an elastic collision, what will be the final velocities of the masses after the collision?

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Mass A (2.0 kg) is moving with an initial velocity of 15 m/s in the + x -direction, and it collides with mass B (5.0 kg), initially at rest. After the collision, the two objects stick together and move as one. What is the change in kinetic energy of the system as a result of the collision?

  1. no change
  2. decrease by 225 J
  3. decrease by 161 J
  4. decrease by 64 J

(c). Because of conservation of momentum, the final velocity of the combined mass must be 4.286 m/s. The initial kinetic energy is ( 0.5 ) ( 2.0 ) ( 15 ) 2 = 225  J . The final kinetic energy is ( 0.5 ) ( 7.0 ) ( 4.286 ) 2 = 64  J , so the difference is −161 J.

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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