Question

Block 1, of mass m1 = 6.70 kg, moves along a frictionless air track with speed v1 = 31.0 m/s. It collides with block 2, of mass m2 = 57.0 kg, which was initially at rest. The blocks stick together after the collision. (Figure 1). The figure shows two states of a system of two blocks, labeled 1 and 2, of masses, m1 and m2, respectively. Block 2 is to the right of block 1. In a state labeled "before collision", block 1 moves to the right with velocity v1, and block 2 is at rest. In a state labeled "after collision", the blocks move together with velocity vf directed to the right. Part A Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically. pi = [?] kg·m/s Part B Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically. vf = [?] m/s Part C What is the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision? Express your answer numerically in joules. ΔK = [?] J

          Block 1, of mass m1 = 6.70 kg, moves along a frictionless air track with speed v1 = 31.0 m/s. It collides with block 2, of mass m2 = 57.0 kg, which was initially at rest. The blocks stick together after the collision. (Figure 1). The figure shows two states of a system of two blocks, labeled 1 and 2, of masses, m1 and m2, respectively. Block 2 is to the right of block 1. In a state labeled "before collision", block 1 moves to the right with velocity v1, and block 2 is at rest. In a state labeled "after collision", the blocks move together with velocity vf directed to the right.

Part A
Find the magnitude pi of the total initial momentum of the two-block system.
Express your answer numerically.
pi = [?] kg·m/s

Part B
Find vf, the magnitude of the final velocity of the two-block system.
Express your answer numerically.
vf = [?] m/s

Part C
What is the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision?
Express your answer numerically in joules.
ΔK = [?] J

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Added by Crystal M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Cal Wilkens

11/04/2022

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Since block 2 is initially at rest, its momentum is zero. Show more…

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Block 1, of mass m1 = 6.70 kg, moves along a frictionless air track with speed v1 = 31.0 m/s. It collides with block 2, of mass m2 = 57.0 kg, which was initially at rest. The blocks stick together after the collision. (Figure 1). The figure shows two states of a system of two blocks, labeled 1 and 2, of masses, m1 and m2, respectively. Block 2 is to the right of block 1. In a state labeled "before collision", block 1 moves to the right with velocity v1, and block 2 is at rest. In a state labeled "after collision", the blocks move together with velocity vf directed to the right. Part A Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically. pi = [?] kg·m/s Part B Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically. vf = [?] m/s Part C What is the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision? Express your answer numerically in joules. ΔK = [?] J

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Transcript

00:01 So we have block 1 of mass 1, which has a mass of 6 .7 kilograms, and it moves along a frictionless air track with speed v1, which is 31 meters per second.

00:21 It then collides with block 2, which has a mass of 57 kilograms, and that block was initially at rest.

00:36 And then after the collision, the blocks stick together.

00:40 So after the collision, let's just draw.

00:44 So before, we know that we have mass 1 going along at some speed v1, and then we have mass 2, which is not moving before the collision.

01:01 And then after the collision, we know that we have mass 1, and mass 2, and they are now stuck together, and they are moving at some speed vf.

01:17 We don't know that speed yet.

01:20 And so for part a, we are asked to find the total initial momentum of the two block systems.

01:30 We want to calculate the momentum before the collision.

01:35 So the initial momentum is equal to mass 1 times v1 plus mass 2 times v2.

01:47 You can plug values in.

01:49 So mass 1 is 6 .7 kilograms times v1, which is 31 meters per second, plus mass 2, which is 57 kilograms, times v2.

02:06 Which is 0 meters per second.

02:09 So this whole part becomes zero.

02:11 And so we get that total momentum before the collision is equal to 2.

02:20 207 .7 kilograms times meters per second.

02:28 And b, we are asked to find that speed vf.

02:35 And so we can do this by relating initial momentum to final momentum.

02:41 So since we already have a value for initial momentum, there's no point in writing this formula up here again.

02:49 But for final momentum, we know that the two masses are now stuck together, so we have mass of the system times vf.

03:03 And so what that gives us is that vf is equal to initial momentum divided by mass of the system...

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