Catching a Ball on Ice

Part A = 6.72 cm/s

Part B = 11.0 cm/s

Solution Below:

Part A

_{f}do Olaf and the ball move afterward?

Express your answer numerically in centimeters per second.

Remember that momentum is conserved, so the momentum before and after Olaf catches the ball will be the same. Before Olaf catches the ball, the momentum is (mass * velocity, or 0.400 * 11.7) 4.68 kg·m/s. This won’t change after Olaf catches the ball, so we can solve for the velocity:

ρ = m * v

4.68 = (0.400 + 69.2) * v

4.68 = 69.6 * v

v = 0.0672 m/s

For some reason, Mastering Physics wants the answer in centimeters per second. So just multiply by 100:

v_{f} = 6.72 cm/s

6.72 cm/s

Part B

Express your answer numerically in centimeters per second.

Remember momentum is conserved. In Part A we saw that the total momentum is (mass * velocity, or 0.400 * 11.7) 4.68 kg·m/s. This will stay the same after the collision. Since the ball will be moving backwards (negative) at 7.40 m/s, it’s momentum after the collision is (7.40 * 0.400) 2.96 kg·m/s. So we can solve for Olaf’s momentum:

ρ_{Olaf} = 4.68 – ρ_{ball}

ρ_{Olaf} = 4.68 – (- 2.96)

ρ_{Olaf} = 7.64

Now we can solve for Olaf’s velocity:

ρ = m * v

7.64 = 69.2 * v

v = 0.110 m/s

For some reason, Mastering Physics wants this in centimeters per second, so just multiply by 100. The answer is 11.0 cm/s

11.0 cm/s