# Inelastic collision and ballistic pendulum

- A 38g bullet is fired at a stationary bag which is hanging vertically from the ceiling. The speed of the bullet is 180m/s. If the mass of the bag is 5.0kg, what is the maximum height it will reach?

- A spring is compress 8.7cm between two carts of 1.0kg and 1.5kg respectively. If the spring constant is 89N/m what is the velocity of the heavier cart.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please see the attachment for a more detailed solution.

When a bullet ( of mass 'm') hits a Block ( of mass M) with a speed 'v' and get arrested in the block, the bullet and block system will start moving with a speed 'Vb'.

This speed can be calculated from the law of conservation of Momentum as

For the qn no 2) :

When a bullet ( of mass 'm') hits a Block ( of mass M) with a speed 'v' and get arrested in the block, the bullet and block system will start moving with a speed 'Vb'.

This speed can be calculated from the law of conservation of Momentum as

Net initial momentum = Net final momentum

( m v + zero ) = ( m + M ) Vb .

; So .

From the given details, Vb=

Because of this speed, the combined system ( bullet and block system) will rise to a height 'h'. To calculate this we need to use the Law of conservation of energy.

According to this,

Kinetic Energy at the bottom = Potential Energy at the highest Point.

Hence the height reached by the system is .

From the given data, the height, reached by the Bag is h = 0.0939 m = 9.39 cm.

For the qn No 3)

In this question, the system is as shown in the figure,

First according to the law of conservation of momentum,

 (1Kg) V1 = ( 1.5 Kg) V2.

 V1 = 1.5 V2 .

 Where V1 is the velocity of the lighter cart and V2 is the velocity of the heavier cart.

Now by using these in the Law of conservation of energy, we have,

 here all quantities were selected in S.I. SYSTEM

solving this we get the V2 as

0.424 ms-1.

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