A 1.50 kg block is attached to a horizontal spring with spring constant 2150 N/m

Question

A 1.50 kg block is attached to a horizontal spring with spring constant 2150 N/m . The block is at rest on a frictionless surface. A 8.00 g bullet is fired into the block, in the face opposite the spring, and sticks. Part A What was the bullet's speed if the subsequent oscillations have an amplitude of 16.6 cm ? A 1.50 kg block is attached to a horizontal spring with spring constant 2150 N/m . The block is at rest on a frictionless surface. A 8.00 g bullet is fired into the block, in the face opposite the spring, and sticks. Part A What was the bullet's speed if the subsequent oscillations have an amplitude of 16.6 cm ? A 1.50 kg block is attached to a horizontal spring with spring constant 2150 N/m . The block is at rest on a frictionless surface. A 8.00 g bullet is fired into the block, in the face opposite the spring, and sticks. Part A What was the bullet's speed if the subsequent oscillations have an amplitude of 16.6 cm ?

Explanation / Answer

The Kinetic energy K of the bullet block must equal the potential energy U of the spring at max amplitude A

So 1/2(M+ m)*V^2 = 1/2*k*A^2

So V = sqrt(k*A^2/(M + m)) = sqrt(2150*0.166^2/(1.50 + 0.008)) = 6.27 m/s

Now from conservation of momentum

m*v = (M + m)*V

so v = (M + m)*V/m = (1.50 + 0.008)*6.27/0.008= 1181.9 m/s

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