A wooden block with m1 = 3kg is sliding across a frictionless surface in the ?x

Question

A wooden block with m1 = 3kg is sliding across a frictionless surface in the ?x direction at 4.7 m/s. A smaller
wooden block with m2 = 0.9 kg block is traveling in the +x direction at 3.2 m/s.

(a) Find the velocity vcm of the center of mass.

(b) What is the velocity of m1 in the COM frame?

What is the velocity of m2 in the COM frame?

Now they make a head-on elastic collision. (This means that in the COM frame, the velocity of each is
reversed.)

What is the velocity of m1 in the COM frame after the collision?

What is the velocity of m2 in the COM frame after the collision?

Transform back into the original frame by adding vcm to the velocity of each block.
What is the velocity of the m1 block in the lab frame after the collision?

7) What is the velocity of the m2 block in the lab frame after the collision?

Explanation / Answer

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

A wooden block with m1 = 2.9kg is sliding across a frictionless surface in the ?x direction at 4.5 m/s. A smaller wooden block with m2 = 1.1 kg block is traveling in the +x direction at 3 m/s. 1) Find the velocity vcm of the center of mass. 2) What is the velocity of m1 in the COM frame? 3) What is the velocity of m2 in the COM frame? 4)Now they make a head-on elastic collision. (This means that in the COM frame, the velocity of each is reversed.) What is the velocity of m1 in the COM frame after the collision? 5)What is the velocity of m2 in the COM frame after the collision? 6)Transform back into the original frame by adding vcm to the velocity of each block. What is the velocity of the m1 block in the lab frame after the collision? 7)What is the velocity of the m2 block in the lab frame after the collision?   

answer

1) vcm = m1v1+m2v2/(m1+m2) = (2.9*(-4.5)+(1.1)(3))/(2.9+1.1) = -2.4375 m/s

2) vel of m1 in com frame = v1 - vcm = -4.5 - (-2.4375) = -2.0625 m/s

3) vel of m2 in com frame = v2 - vcm = 3 - (-2.4375) = 5.4375 m/s

4) vm1 after collision in com frame vm1'= 2.0625 m/s (since velocity reverses in com frame)

5) vm2 after collision in com frame = -5.4375 m/s

6) vm1 after colliion = vm1'+vcm = 2.0625 + (-2.4375) = -0.375 m/s

7) vm2 after collision = -5.4375 + (-2.4375) = -7.875 m/s

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