Question

A bullet of mass 100 g moving with a velocity of 250 m/s hits a block of mass 2 kg resting on a smooth table. It comes out with a speed of 150 m/s. The block moves with a velocity of __ m/s.

1

Answer 1

Given :

• Mass of the bullet = 100 g.

• Intial velocity of the bullet = 250 m/s.

• Mass of the block = 2 kg.

• Final Velocity of the bullet = 150 m/s.

To find :

Final Velocity of the block.

Solution :

We know the Law of conservation of momentum i.e,

∴ total momentum before collision = total momentum after collision.

Mathematically,

$\boxed{\bf{m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{1}v_{2}}}$

Where :

• m = Mass
• u = Intial velocity
• v = Final Velocity

Now using the formula for Conservation of Momentum and substituting the values in it, we get :

$:\implies \bf{m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}}$

Here, the initial velocity of the block will be 0 since it's starting from rest position.

$:\implies \bf{m_{1}u_{1} + m_{2}(0) = m_{1}v_{2} + m_{1}v_{2}}$

$:\implies \bf{m_{1}u_{1} + 0 = m_{1}v_{1} + m_{2}v_{2}}$

$:\implies \bf{m_{1}u_{1} = m_{1}v_{1} + m_{2}v_{2}}$

$:\implies \bf{0.1 \times 250 = 0.1 \times 150 + 2 \times v_{2}}$

$:\implies \bf{25 = 15 + 2 \times v_{2}}$

$:\implies \bf{25 - 15 = 2 \times v_{2}}$

$:\implies \bf{10 = 2 \times v_{2}}$

$:\implies \bf{\dfrac{10}{2} = v_{2}}$

$:\implies \bf{5 = v_{2}}$

$\boxed{\therefore \bf{v_{2} = 5\:ms^{-1}}}$

Hence the final Velocity of the block is 5 m/s.

Answer 2

Given :  A bullet of mass 100 g moving with a velocity of 250 m/s hits a block of mass 2 kg resting on a smooth table. It comes out with a speed of 150 m/s.

To Find : The block moves with a velocity of __ m/s.

Solution:

Conservation of momentum :

Initial Momentum

= (100/1000) * 250  + 2 * 0

= 25  kg m/s

Final Momentum  = 25 kg m/s

2 * v +  (100/1000)150  = 25

=> 2v  + 15 = 25

=> 2v = 10

=> v = 5

The block moves with a velocity of 5  m/s.  

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