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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

Respuesta :

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

What is momentum conservation principle?

When there is no external force acting on the system, the momentum remains conserved.

For inelastic collision, after collision both objects travel with common speed.

m1u1 + m2u2 =(m1 +m2)v

Substitute initial velocity of bullet u1 =320 m/s , initial velocity of block u2 =0, mass of bullet m1 = 0.1 kg and mass of block m2 = 14.9 kg.

Solve for the final velocity of bullet,

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

v = 2.133 m/s

Thus, the  block with the bullet lodged in block now travelling at 2.133 m/s.

Learn more about momentum conservation principle.

https://brainly.com/question/14033058

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