A 0.0240 kg bullet moving horizontally at 500 m/s embeds itself into an initiall
ID: 1282989 • Letter: A
Question
A 0.0240 kg bullet moving horizontally at 500 m/s embeds itself into an initially stationary 0.500 kg block.
(a) What is their velocity just after the collision? ___ m/s
(b) The bullet-embedded block slides 8.0 m on a horizontal surface with a 0.30 kinetic coefficient of friction. Now what is its velocity? ___ m
(c) The bullet-embedded block now strikes and sticks to a stationary 2.00 kg block. How far does this combination travel before stopping? ___ m
I'd really appreciate it if you could explain how to solve this. Thank you!
Explanation / Answer
a)
The momentum of the bullet before the collision is 0.0240 * 500 = 12 kg m/s.
The momentum of the block is 0 since it's stationary (so v = 0).
The total momentum after the collision must be the same, i.e. 12 kg m/s. The bullet has embedded itself in the block so they can be treated as one object. That means 0.5240 * v = 12, so v = 12 / 0.5240 = 22.90 m/s.
b)
The frictional force is the coefficient of friction multiplied by the force going towards the surface, which in this case will have a magnitude equal to the force of gravity. So 0.3 * 0.5240 * 9.8 = 1.541 N, acting in the opposite direction to the block's movement. The acceleration of the block is then F / m = -1.541 / 0.5240 = -2.941 m/s^2.
If acceleration (a) is constant and in the opposite direction to the velocity then v^2 = u^2 - 2as where v = final velocity, u = initial velocity, s = displacement. So v^2 = 22.90^2 - 2 * 2.941 * 8 = 477.354, so v = 21.85 m/s.
c)
The block with the bullet now has a momentum of 0.5240 * 21.85 = 11.45 kg m/s. The new block has no momentum like before. So the momentum of the combined blocks after the collision must equal 11.45 kg m/s too. That means 2.5240 * v = 11.45, so v = 4.54 m/s.
The acceleration is the same as before (mass cancels out), and when the blocks stop v will = 0. So 4.54^2 - 2 * 2.941 * s = 0, so 5.882 * s = 20.6116, so s = 20.6116/5.882 = 3.50 m.
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