challenging physics?

Question

block 1 (mass 1.8 kg) is moving rightward at 12 m/s and block 2 (mass 4.6 kg) is moving rightward at 2.8 m/s. The surface is frictionless, and a spring with spring constant of 1160 N/m is fixed to block 2. When the blocks collide, the compression of the spring is maximum at the instant the blocks have the same velocity. Find the maximum compression.


A 5.20 g bullet moving at 620 m/s strikes a 670 g wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to 428 m/s.
(a) What is the resulting speed of the block?
m/s
(b) What is the speed of the bullet-block center of mass?
m/s

Answer 1

Ok, I am going to do the second problem first ok ?, is the easier one.

b ) the linear momentum will be the same before and after the strike :

5.20*620 = 5.20*428 + 670*V

5.20*(192) = 670*V

V =1.5 m/s >>> speed of the wooden block

to calculate the speed of the bullet - block center of mass after the strike, you must use the speed of the wooden block : 1.5 m/s, so then :

V of center of mass = 5.20*428 + 1.5*670 / 670 + 428

V of center of mass = 3230.6 / 1098 = 2.94 m/s >>> thats the speed of the center of mass.

For the first question : I have a little doubt, but, let's say that :

The total kinetic energy will before the strike, will be transform into potential energy for the spring, then :

1.8*12^2 / 2 + 4.6*2.8^2 / 2 = K*x^2 / 2

where K is the spring constant

and there is no kinetic energy after or when they strike, because, the maximum compression happens when the two blocks stop.

K = 1160 N / m

1160*x^2 / 2 = 147.6

then x = 0.5 m = 50 cm

Answer 2

This looks like a homework question. Which part do you not understand, so that we can offer you help?