A bullet with a mass of 0.1 kg is travelling at 320 m/s when it strikes a stationary block used for target practice. The block has a mass of 14.9 kg. The bullet, now embedded in the block, causes the block to fall off the post that it was resting upon. How fast is the block with the bullet lodged in it now travelling?

Velocity: _ m/s

Respuesta :

The block with the bullet after collision lodged in it is now travelling at 2.133 m/s.

What is conservation of momentum principle?

When there is no external force acting on the system , the initial momentum is equal to final momentum.

The momentum is the product of mass and its velocity.

Given is the mass of bullet m₁ = 0.1 kg, mass of block m₂ =14.9kg, velocity of bullet before collision u₁ =320 m/s, velocity of block before collision u₂ =0. The final velocity of block and bullet both is v.

According to the momentum conservation principle,

m₁u₁ +m₂u₂ = (m₁ +m₂) v

Substitute the values into the expression to find the final velocity,

0.1 x 320 +14.9 x 0 =(0.1 +14.9) v

v = 2.133 m/s

Thus, the block with the bullet logged in it is moving at 2.133 m/s.

Learn more about conservation of momentum principle.

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