A rifle bullet with mass 9.0 grams strikes and embeds itself into a block with m

Question

A rifle bullet with mass 9.0 grams strikes and embeds itself into a block with mass 992 grams that rests on a frictionless, horizontal surface. The block is attached to a spring. The impact causes the spring to compress 15.0 cm before stopping. A previous calibration of the spring found that it takes a force of 0.750 Newtons to compress the spring 0.25 cm Find the magnitude of the block's velocity immediately after impact What was the initial speed of the bullet? What was the impulse of the bullet on the block took 0.05 seconds for the bullet to embed itself into block?

Explanation / Answer

The spring constant of the spring is given by k = F/x = 0.75N/0.0025m = 300 N/m.

As the block moves on a frictionless surface, all of the kinetic energy from the bullet is transferred first to kinetic energy in the block, then to strain energy in the spring.

The strain energy is given by E = 0.5*k*x^2 = 3.375 J
The kinetic energy of the combined block and bullet E = 0.5*(m+M)*v^2 = 3.375 J
v = 1.84 m/s
where mass = m+M = 1.001 kg (0.992+0.009kg)

b) Vo = (m+M)v/m = 1.001*1.84/0.009 = 204.65 m/s

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