Please show all steps on how to get to the final answer please! An 8.0g bullet i

Question

Please show all steps on how to get to the final answer please!

An 8.0g bullet is shot into a 4.0kg block, itself initially at rest on a frictionless horizontal surface. The bullet remains lodged in the block. The block and bullet then move together into an ideal massless spring and compress it by 8.7cm (at which point the block and bullet momentarily come to rest). The spring constant of the spring is 2400N/m. The initial velocity of the bullet is closest to? 1070m/s 1120m/s 960m/s 1200m/s 1010m/s

Explanation / Answer

kinetic energy of block-bullet system converts into spring PE.

using energy conservation,

kx^2 / 2 = m v^2 / 2

2400 (0.087^2) = (4 + 0.008) v^2

v = 2.13 m/s

now Applying momentum conservation for collision,

0.008 v0 + 4x0 = 4.008x 2.13

v0 = 1066.6 m/s

Ans(A)

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