2008: Problem 22

Topic: Collisions
Concepts: Ballistic pendulum, Circular motion, Forces in mechanics

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Solution:

By conservation of linear momentum,

$$m_1v_0=(m_1+m_2)v_b$$

$$v_b=\frac{m_1}{m_1+m_2}v_0$$

We conserve energy between the bottom and top of the vertical circle,

$$\frac{1}{2}(m_1+m_2)v_b^2=\frac{1}{2}(m_1+m_2)v_t^2+2(m_1+m_2)gL$$

$$v_b^2=v_t^2+4gL$$

The minimum velocity at the top of the vertical circle is attained when the tension in the string goes to zero. Only gravity provides the centripetal acceleration so

$$mg=\frac{mv_t^2}{L}$$

$$v_t^2=gL$$

Substituting this into the equation for $v_b$,

$$v_b^2=5gL$$

$$\frac{m_1}{m_1+m_2}v_0=\sqrt{5gL}$$

$$v_0=\frac{(m_1+m_2)\sqrt{5gL}}{m_1}$$

so the answer is E.

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