A0.0240 kg bullet moving horizontally at 400 m/s embeds itself into an initially stationary 0.500 kg block. (a) what is their velocity (in m/s) 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 (in m/s)? m/s (c) the bullet-embedded block now strikes and sticks to a stationary 2.00 kg block. how far (in m) does this combination travel before stopping? m

(a) 18.3 m/s

According to the law of conservation of momentum, the total initial momentum of the system must be equal to the total final momentum, so we have

where

m = 0.0240 kg is the mass of the bullet

u = 400 m/s is the initial speed of the bullet

M = 0.5 kg is the mass of the block

v is the final speed of the block+bullet together

Solving for v, we find the velocity after the collision

(b) 17.0 m/s

The frictional force acting on the bullet-block system is

where

is the coefficient of kinetic friction

The acceleration due to the frictional force, therefore, will be equal to the frictional force divided by the total mass:

The system travels for a distance of

d = 8.0 m

So we can find the final velocity using the equation:

where

v = 18.3 m/s is the initial velocity, found at point a). Substituting,

(c) 2.1 m

We can use again the law of conservation of momentum:

where

v = 17.0 m/s is the initial velocity of the initial bullet+block system

M = 2.00 kg is the mass of the second block

v' is the final velocity of the system

Solving for v',

The acceleration of the system on the rough surface is still

So we can find the distance covered by using again the formula used before, and requiring that the final velocity should be zero (v''=0):

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