First slide

Types of collision

Question

A ball of mass m moving with a horizontal velocity v strikes the bob of a pendulum at rest. Mass of the bob of the pendulum is also m. During this collision, the ball sticks with the bob of the pendulum. The height to which the combined mass rises will be (g = acceleration due to gravity)

Moderate

A

$\frac{v^{2}}{4 g}$
B

$\frac{v^{2}}{8 g}$
C

$\frac{v^{2}}{g}$
D

$\frac{v^{2}}{2 g}$

Solution

From the law of conservation of momentum,
   $mv + m \times 0 = (m + m) v'$
$\Rightarrow$    $v' = \frac{mv}{(m + m)} \Rightarrow v' = \frac{v}{2}$
and we know that, $h= \frac{u^{2}}{2 g}$ $(∵ v^{2} - u^{2} = 2 gh)$
Here,    $u = v'$
So, $h = \frac{{(v')}^{2}}{2 g}$    $(∵ v' = \frac{v}{2})$
$\Rightarrow$ $h = \frac{v^{2}}{8 g}$

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