First slide

Second law

Question

A bullet moving with a velocity of 30 $\sqrt{2}$ ms^-1 is fired into a fixed target. It penetrated into the target to the extent of s meter. If the same bullet is fired into a target of thickness $\frac{s}{2}$ metres and of the same material with the same velocity, then the bullet comes out of the target with velocity, is [EAMCET 2014]

Moderate

A

20 ms^-1
B

30 ms^-1
C

20 $\sqrt{2}$ ms^-1
D

10 $\sqrt{2}$ ms^-1

Solution

Given, u = 30 $\sqrt{2}$ ms^-1
Distance = s meter
Let a be the acceleration of the bullet.
According to the first condition, (v = 0)
From third equation of motion, $v^{2} = u^{2} - 2 a s$
$0 = 900 \times 2 - 2 a s \Rightarrow a = \frac{900}{s} {ms}^{- 2}$
For second condition,
$v^{2} = u^{2} - 2 a s = (30 \sqrt{2})^{2} - 2 \times \frac{900}{s} \times \frac{s}{2}$
$\Rightarrow v^{2} = 1800 - 900$
$\Rightarrow v = \sqrt{900} = 30 {ms}^{- 1}$

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