A 5.00-g bullet moving with an initial speed of v i = 460 m/s is fired into and

Question

A 5.00-g bullet moving with an initial speed of vi = 460 m/s is fired into and passes through a 1.00-kg block as shown in the figure below. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 890 N/m. The block moves d = 4.20 cm to the right after impact before being brought to rest by the spring.

A 5.00-g bullet moving with an initial speed of vi = 460 m/s is fired into and passes through a 1.00-kg block as shown in the figure below. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 890N/m. The block moves d = 4.20 cm to the right after impact before being brought to rest by the spring.

Explanation / Answer

Energy of the system is conserved hence

0.5mvi^2 = 0.5kx^2 + 0.5mvf^2 (where vf is the speed of the bullet with which it emerges)

0.5 * 0.005 * 460^2 = 0.5 * 890 * (4.2 * 10^-2)^2 + (0.5 * 0.005 * vf^2)

vf = sqrt(((0.5 * 0.005 * 460^2) - (0.5 * 890 * (4.2 * 10^-2)^2))/(0.5 * 0.005)) = 459.658 m/s

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