Question

Select One Option Correct from the following :A ball of mass 0.2 kg rests on vertical post of height 5 m. A bullet of mass 0.01 kg, travelling with velocity V m/s in horizontal direction, hits the centre of the ball. After the collision, the ball and the bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The initial velocity V of the bullet is [Take $g=10 m/s^{2}$.]
(A)250 m/s
(B)$250\sqrt{2} m/s$
(C)400 m/s
(D)500 m/s

Answer 1

d option is correct I think

Answer 2

Answer:

500m/s

Explanation:

Solve this problem with conservation of momentum, p = mv  

The momentum of the bullet p1 = 0.01V is conserved as: p1 = p2+p3  

p1 = p2+p3  

0.01V = 0.01v1+0.2v2  

V = (0.01v1+0.2v2)/0.01  

Find v1 and v2, the velocities after the collision.  

Gravitational acceleration g = 9.81 m/s^2 makes both objects fall a height of 5m  

in the same amount of time, and both objects has no initial vertical velocity.  

y = y0+v0t+gt^2/2  

5 = 0+0t+4.905t^2  

t = √(4.905)5) = 1.01 s  

The bullet travels 100 m in 1.01 s and the ball travels 20 m in the same amount of time.  

To find their initial horizontal velocities right after the collision:  

Bullet: v1 = x/t = 100/1.01 = 99.0 m/s  

Ball: v2 = x/t = 20/1.01 = 19.8 m/s  

==> V = (0.01*99.0+0.2*19.8)/0.01 = 495 ≈ 500 m/s