# physics questions!! Answer them and will rate lifesaver!!?

A 1500 kg car with a velocity of 15 m/s is traveling behind a 1000 kg car with a velocity of 10 m/s. If the velocity of the 1500 kg car is 12.5 m/s after hitting the car, what is the velocity of the other car?

A 30 gram bullet at 200 m/s is fired horizontally into a 1.25 kg wood block at rest. How far does it slide if the coefficient of friction is 0.3?

Relevance

Use conservation of momentum:

The initial momentum of the larger car is:

p1i = m1*v1i = (1500 kg)(15 m/s) = 22500 kg*m/s

The initial momentum of the smaller car is:

p2i = m2*v2i = (1000 kg)(10 m/s) = 10000 kg*m/s

The final momentum of the larger car is:

p1f = m1*v1f = (1500 kg)(12.5 m/s) = 18750 kg*m/s

Using conservation of momentum, the final momentum of the smaller car is:

p1f + p2f = p1i + p2i

p2f = p1i + p2i - p1f = (22500 kg*m/s) + (10000 kg*m/s) - (18750 kg*m/s) = 13750 kg*m/s

Thus, the final velocity of the smaller car is:

v2f = p2f/m2 = (13750 kg*m/s)/(1000 kg) = 13.75 m/s

The total kinetic energy of the bullet is:

K = (1/2)(b)v^2

The weight of the block and the bullet is:

mg = (B+b)g

Since the only other vertical force is the normal force:

N = mg = (B+b)g

So, the force of friction is:

F = u*N = u*(B+b)*g

The work done by friction is:

W = F*d = u*(B+b)*g*d

Since there is no kinetic energy after the block comes to rest, the work done by friction is equal to the energy of the bullet. So:

W = K

u*(B+b)*g*d = (1/2)(b)v^2

d = [(1/2)(b)v^2]/[u*(B+b)*g] = [(1/2)(0.030 kg)(200 m/s)^2]/[(0.3)*(1.25 kg + 0.030 kg)(9.81 m/s^2)]

d = 159.3 m