Question

1. During an ice-skating performance, an initially motionless 80.0-kg clown throws a fake barbell away. The clown’s ice skates allow her to recoil frictionlessly. If the clown recoils with a velocity of 0.500 m/s and the barbell are thrown with a velocity of 10.0 m/s, what is the mass of the barbell?
2. How much kinetic energy is gained by this maneuver?

Answer 1

a) Since the ice is frictionless, we can use the momentum conservation law:

$$p = p_1 + p_2$$

where $p = 0$ is the momentum of the clown and barbell before she throws it away (since they are motionless), $p_1, p_2$ are their momentums after the throw respectively. By definition, the momentums are:

$$p_1 = m_1v_1\\ p_2 = m_2v_2$$

where $m_1 = 80kg$ is the mass, and $v_1 = 0.5m/s$ is the speed of the clown after the throw; $m_2$ is the mass, and $v_2 = 10m/s$ is the speed of the barbell after the throw.

Thus, obtain:

$$0 = m_1v_1+m_2v_2\\ m_2 = \left| -\dfrac{m_1v_1}{v_2} \right|$$

The module here arises due to fact that the speeds $v_1,v_2$ actually have different sign, since they are directed in opposite directions. Thus, obtain:

$$m_2 = \dfrac{80kg\cdot 0.5m/s}{10m/s} = 4kg$$

b) The total kinetic energy is:

$$K = \dfrac{m_1v_1^2}{2} + \dfrac{m_2v_2^2}{2}\\ K = \dfrac{80\cdot 0.5^2}{2} + \dfrac{4\cdot 10^2}{2} = 210J$$

Answer. a) 4kg, b) 210 J.