Question

From the top of a tower of height 100 m a 10 gm block is dropped freely and a 6gm bullet is fired vertically upwards from the foot of the tower with velocity 100m * s ^ - 1 simultaneously. They collide and stick together. The common velocity after collision is (g = 10m * s ^ - 2) 1) 27.5 ms^ -1 2) 150 ms^ -1 3) ms^ -1 3)40 ms^ -1 4)100 ms^ -1​

Answer 1

Answer:

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Explanation:

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Answer 2

Given:

Height from which the block is dropped = 100 m

Mass of the block = 10 gm

Mass of the bullet = 6 gm

Velocity of the bullet = 100 m/s

Acceleration due to gravity = 10 m/s²

To find:

The common velocity after collision.

Solution:

We are considering downward direction to be positive.

Their relative velocity is vr  =100m/s and relative acceleration ar  =0  

If they collide at t seconds later then:

Sr = vr*t + 1/2*ar*t²

100 = 100t + 0

t = 1 sec

Therefore after 1 sec,

V block = g*t = 10 m/s

V bullet = 100 - g*t = 90 m/s

Now by conservation of momentum we get:

m block * v block + m bullet * v bullet = m of both* v of both

0.01* 10 + 0.006* 90 = (0.01 + 0.006) * V of both

V = -27.5 m/s

Therefore the velocity of the system after collision will be 27.5 m/s in the upward direction.