Question

A bullet of mass 20g moving with a velocity of 200m/s strikes and get embedded in to aa stationaary wooden block of mass 980g .find velocity with which the block moves.

Answer 1

Explanation:

It is given that,

Mass of a bullet, m₁ = 20 g = 0.02 kg

Speed of a bullet, v₁ = 200 m/s

The bullet strikes and get embedded in to aa stationary wooden block of mass 980g. It means that, m₂ = 980 g + 20 g = 1000 g = 1 kg

Let v₂ is the velocity with which the block moves. The momentum will remain conserved. So, using the conservation of momentum as :

$m_1v_1=m_2v_2\\\\v_2=\dfrac{m_1v_1}{m_2}\\\\v_2=\dfrac{0.02\times 200}{1}\\\\v_2=4\ m/s$

So, the block will move with a velocity of 4 m/s. Hence, this is the required solution.

Answer 2

Here it is given that,

Mass of bullet $M_{1}$ is $20$g = $0.02$ kg

Mass of wooden $M_2$ is $980$g = $0.98$ kg

For bullet initial velocity $u_1$ is $200$ m/s

We have to determine the velocity at which the block moves.

For wooden block initial velocity $u_2$ is $0$ m/s

The block and the bullet both move with the same velocity after collision.

Let, the velocity be 'v',

By the Law of Momentum Conservation,

$m_1u_1+m_2u_2=m_1v+m_2v$

Therefore,

$=>0.02\times 200+0\times 0\times 0.98=v(m_1+m_2)$

$=>4=v(0.98+0.02)$

$=>4=v(1)$

$=>v=4$

Hence, the velocity at which the block moves is $4$ m/s