## Mastering Physics Solutions: Exercise 6.67

C Exercise 6.82

Part A = 0.10m

Solutions Below:

A 1.0 kg object moving at 1.1 m/s collides elastically with a stationary 1.0 kg object, similar to the situation shown in the figure.

Part A

How far will the initially stationary object travel along a 37° inclined plane? (Neglect friction.)

This has to be one of the worst Mastering Physics problems I’ve ever seen. The diagram for this problem is wrong (they even admit that in the description. I don’t know if MP was lazy or purposely trying to mess people up (lame), but I’m not going to even put the picture in here. I’ll just explain the problem to you so that it’s clear.

What we have is 2 blocks, both with a mass of 1kg. The block on the left is sliding at 1.1m/s towards the block on the right, and behind the block on the right is a ramp with an angle of 37°. So when the left block hits the right block, it’s going to push it up the ramp. You have to find the distance the second block moves.

Since this is an elastic collision, energy is conserved. So the kinetic energy is given by:

KE = 1/2mv2
KE = 1/2(1)(1.1)2
KE = 0.605J

When the first block hits the second, all of this energy will be put into the second block (since the collision is said to be elastic, for this problem we can assume it’s perfectly elastic). Since the mass of the second block is the same, it will end up with the same velocity (the first block will come to a stop). I only put the above equation so you could follow the logic.

So the second block will end up with a velocity of 1.1m/s.

Now all we have to do is solve the kinematic equation for distance:

vf2 = v02 + 2a(d)

At the maximum height, gravity will have slowed the block’s velocity to zero. Since the ramp is at an incline, we need to solve for ‘g’ (since the ramp isn’t straight up, not all of the 9.8m/s2 will act on the block):

sin(37) * -9.8 = -5.9m/s2

So:

vf2 = v02 + 2a(d)
0 = 1.12 – 2(5.9)(d)
1.21 = 11.8d

d = 0.10m (2 significant figures)

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