A 0.3 kg block compresses a spring of spring constant 1300 N/m by 0.12 m. After

Question

A 0.3 kg block compresses a spring of spring constant 1300 N/m by 0.12 m. After being released from rest, the block slides along a smooth, horizontal and frictionless surface before colliding elastically with a 1 kg block which is at rest. (Assume the initial direction of motion of the sliding block before the collision is positive.)

A) What is the velocity of the 0.3kg block just before striking the 1 kg block?

B) What is the velocity of the 1 kg block after the collision?

C) What is the velocity of the 0.3 kg block after the collision?

D)

Explanation / Answer

(a)1/2mv^2 = 1/2kx^2 ===> v^2 = kx^2/m ===> v = 7.9 m/s (b) solving M1v = M1u1 + M2u2 and M1v^2/2 = M1u1^2/2 + M2u2^2/2 we get u2 = v(2M1)/(M1+M2) = 3.646 m/s (c) u1 = v(M1 - M2)/(M1+M2) = - 4.254 m/s (d)1/2mv^2 = 1/2kx^2 ===> x^2 = mv^2/k = 0.065 m

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