# Simple Harmonic motion and collisions

A block of mass 2kg oscillates on the end of a spring in SHM with a period of 20ms. The position of the block is given by: x=(1.0cm)cos(wt+2pi). Another block of mass 4kg slides toward the oscillating block with velocity of 6m/s. The two blocks undergo a completely inelatic collision at time t=5.0ms.

a)what is the amplitude of oscllaltion after collision?
I know momentum is conserved in the collision and time t=.005 secons, the velocity of the oscillating block is O. So I got V(final)=4m/s.
But I am not sure how to account for this with the spring force to get the final amplitude.

b) What is the frequency of the resulting SHM?
without A, I dont think I can do this part, so I havent tried yet...

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cherylsc said:
A block of mass 2kg oscillates on the end of a spring in SHM with a period of 20ms. The position of the block is given by: x=(1.0cm)cos(wt+2pi).
Are you sure it's cos(wt+2pi)? Might as well make it cos(wt).

Another block of mass 4kg slides toward the oscillating block with velocity of 6m/s. The two blocks undergo a completely inelatic collision at time t=5.0ms.

a)what is the amplitude of oscllaltion after collision?
I know momentum is conserved in the collision and time t=.005 secons, the velocity of the oscillating block is O. So I got V(final)=4m/s.
If the position is given by:
x=(1.0cm)cos(wt+2pi) then, t=5.0ms corresponds to an angle of pi/2. The velocity equation is:
v=-w(1.0cm)sin(wt+2pi). v does not equal 0 at t=5.0ms.

I'd do part b first to get the new w... then use the fact that the magnitude of the maximum velocity = wA. Since you have the maximum velocity (you get it from the conservation of momentum etc...) you can get A (amplitude).

But I am not sure how to account for this with the spring force to get the final amplitude.

b) What is the frequency of the resulting SHM?
without A, I dont think I can do this part, so I havent tried yet...
w=sqrt(k/m).

You can calculate k, then solve for the new w using a mass of 6kg

oops

I meant (1/2)pi

thanks