First slide

Kinetic energy

Question

A bullet of mas m moving horizontally with velocity v hits a block of wood of mass M, resting on a smooth horizontal plane. The fraction of energy of the bullet dissipated in the collision it self is (assume collision to be inelastic)

Moderate

A

$\frac{mv}{m + M}$
B

$\frac{m}{M + m}$
C

$\frac{Mm}{m + M}$
D

$\frac{M}{m + M}$

Solution

Applying the law of conservation of momentum, we have
$mV = (m + M) V V = (\frac{mv}{m + M}) Loss of K.E. = \frac{1}{2} {mv}^{2} - \frac{1}{2} (m + M) V^{2} = \frac{1}{2} {mv}^{2} - \frac{1}{2} (m + M) {(\frac{mv}{M + m})}^{2} = \frac{1}{2} {mv}^{2} [1 - \frac{m}{m + M}] = \frac{1}{2} {mv}^{2} [\frac{M}{m + M}] Fraction of K.E. dissipated = \frac{Loss of K.E.}{Initial K.E.} = (\frac{M}{m + M})$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App