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Bouncing Ball Example

The first col...

Question

The first collision takes place at time $t_{1}$ and the second collision takes place at time $t_{2}$ . Find $t_{2} - t_{1}$ :

\frac{2 L}{v_{0}}

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\frac{L}{v_{0}}

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\frac{L}{2 v_{0}}

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\frac{L}{3 v_{0}}

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Solution

The correct option is B $\frac{L}{v_{0}}$

Let the velocities of the cart and the bead just after the first collision are $V_{c}$ and $V_{b}$ respectively. Momentum is conserved just before and just after the collision:

${M V}_{0} = M V_{c} + m V_{b} . . . . . . (1) (- V_{0}) e = V_{b} - V_{c} (C o e f f i c i e n t o f r e s t i t u t i o n) \Rightarrow V_{o} = V_{c} - V_{b} (s i n c e e = 1) . . . . . . (2) \Rightarrow V_{c} = \frac{(M - m)}{(M + m)} V_{0}$

$\Rightarrow V_{b} = \frac{2 M}{(M + m)} V_{0}$
with respect to the cart the ball has to move a distance of L for the second collision. time between two collisions is
$t = \frac{L}{V_{\frac{b}{c}}} = \frac{L}{\frac{2 M}{(M + m)} V_{0} - \frac{(M - m)}{(M + m)} V_{0}} = \frac{L}{V_{0}}$

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The first collision takes place at time t1 and the second collision takes place at time t2. Find t2−t1 :

The first collision takes place at time $t_{1}$ and the second collision takes place at time $t_{2}$ . Find $t_{2} - t_{1}$ :