Question

A small small bullet of mass m is fired with velocity uat an angle theta with the horizontal, freely under gravity. at the highest point of its path, it strikes with a wooden block (mass M) close and emerges out with half of its velocity(just before collision) close. during the impact (collision) close, the string (supporting the wooden block) close remain almost vertical. (given M >m) .​

Answer 1

Answer:

your answer dear........... option c

Answer 2

Given:

A small bullet of mass 'm' is fired with velocity 'u' at an angle 'θ' with the horizontal, freely under gravity.

To find:

final velocity =?

Step to step explanation:

A bullet, after firing strikes the wooden block.

Let

m =  mass of the bullet before the strike

M = mass of the wooden block

u = initial velocity of the bullet before firing

v = final velocity of the bullet after it strikes the wooden block

The total energy of the bullet is conserved.

i.e. total energy before a strike is equal to total energy after a strike.

$\displaystyle \frac{1}{2} m (vcos\theta)^2 + \frac{1}{2}M(0)^2 = \frac{1}{2}mv^2 + \frac{\frac{1}{2} M (vcos\theta)^2}{2}$

$\displaystyle \frac{m ( vcos\theta)^2 }{M^2} = \frac{1}{2}Mv^2$

$\displaystyle v= \sqrt{ \frac{m}{M} vcos\theta} ^2$

v = $\displaysyle { \frac{m}{M} vcos\theta }$

Option C is the correct answer.