Simple Harmonic motion and collisions

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A 2kg block oscillates on a spring with a period of 20ms, described by the equation x=(1.0cm)cos(wt+2pi). A 4kg block moving at 6m/s collides inelastically with the oscillating block at t=5.0ms. Momentum conservation is crucial for determining the final velocity post-collision, which is calculated to be 4m/s. To find the new amplitude after the collision, the relationship between maximum velocity and amplitude must be utilized, requiring the calculation of the new angular frequency. The resulting frequency of the new simple harmonic motion can be derived once the amplitude is known.
cherylsc
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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 don't think I can do this part, so I haven't 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 don't think I can do this part, so I haven't 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
 
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