A bullet with mass 15 grams approaches and imbeds in a target wooden block that

Question

A bullet with mass 15 grams approaches and imbeds in a target wooden block that has a mass of 5.000 kg and is initially at rest. After the collision, the block (with the bullet) still imbedded in it. slides along a rough surface 40 cm before stopping. If the coefficient of friction between the block and the surface is 0.20. determine the initial velocity of the bullet. There are two parts to this problem, the initial "collision" between the bullet and block, and the sliding part. In the first, we can use conservation of momentum. In the second, we can use our formulas for forces and kinematics, assuming constant deceleration. Because the two objects "stick together" or more realistically stated, the bullet is imbedded in the block, the right hand side can be simplified. Both objects have the same final velocity! This is called a completely inelastic collision. Rewrite your equation using this idea.

Explanation / Answer

friction force on block.

f = uk N = uk m g

a = - f / m = - uk g = - 0.20g

Applying vf^2 - vi^2 = 2 a d

0^2 - v^2 = 2(-0.20g)(0.40)

v = 1.25 m/s ....This is speed of block just after bullet imbedded in it.

now applying momentum conservation to find the initial speed of bullet.

0.015u + 5 x 0 = (5 + 0.015)v

u = (5.015 x 1.25) / (0.015) = 418 m/s

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