Question

A body of mass m1 standing on a smooth plane of ice throws a ball of mass m2 horizontally on the surface of ice. If x be the distance between them after a time t, then find the amount of work done by the boy in throwing the ball?

Answer 1

Answer on Question #48487-Physics-Mechanics-Kinematics-Dynamics

A body of mass $m_{1}$ standing on a smooth plane of ice throws a ball of mass $m_{2}$ horizontally on the surface of ice. If $x$ be the distance between them after a time $t$ , then find the amount of work done by the boy in throwing the ball?

Solution

The work done by the boy in throwing the ball is equal to the difference of kinetic energy of the system:

$$W = K _ {f} - K _ {i} = \left(\frac {m _ {1} v _ {1} ^ {2}}{2} + \frac {m _ {2} v _ {2} ^ {2}}{2}\right) - 0 = \frac {m _ {1} v _ {1} ^ {2}}{2} + \frac {m _ {2} v _ {2} ^ {2}}{2}.$$

We know that

$$v _ {1} + v _ {2} = \frac {x}{t}.$$

From the conservation of momentum:

$$m _ {1} v _ {1} = m _ {2} v _ {2} \rightarrow v _ {2} = v _ {1} \frac {m _ {1}}{m _ {2}}.$$

Thus

$$v _ {1} + v _ {1} \frac {m _ {1}}{m _ {2}} = \frac {x}{t} \rightarrow v _ {1} = \frac {m _ {2}}{m _ {1} + m _ {2}} \frac {x}{t}; v _ {2} = \frac {m _ {1}}{m _ {1} + m _ {2}} \frac {x}{t}.$$

So,

$$W = \frac {m _ {1}}{2} \left(\frac {m _ {2}}{m _ {1} + m _ {2}} \frac {x}{t}\right) ^ {2} + \frac {m _ {2}}{2} \left(\frac {m _ {1}}{m _ {1} + m _ {2}} \frac {x}{t}\right) ^ {2} = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \frac {x ^ {2}}{2 t ^ {2}}.$$

Answer: $\frac{m_1 m_2}{m_1 + m_2} \frac{x^2}{2t^2}$ .

http://www.AssignmentExpert.com/