A 0.3 kg block compresses a spring of spring constant 1900 N/m by 9×10 2 m. Afte

Question

A 0.3 kg block compresses a spring of spring constant 1900 N/m by 9×102 m. After being released from rest, the block slides along a smooth, horizontal and frictionless surface before colliding elastically with a 1.3 kg block which is at rest. (Assume the initial direction of motion of the sliding block before the collision is positive.)

Part A

What is the velocity of the 0.3 kg block just before striking the 1.3 kg block?

Part B

What is the velocity of the 1.3 kg block after the collision?

Part C

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

Part D

If the 0.3 kg block comes back into contact with the spring, what is the maximum compression of the spring? (If the 0.3 kg block does NOT come back into contact with the spring, put zero here.)

Explanation / Answer

A) Apply energy conservaion

0.5*m*u^2 = 0.5*k*x^2

u = x*sqrt(k/m)

= 9*10^-2*sqrt(1900/0.3)

= 7.16 m/s <<<<<<<---------Answer

b) let m1 = 0.3 kg, u1 = 7.16 m/s

m2 = 1.3 kg, u2 = 0

let v1 and v2 are the velocities of m1 and m2 after the collsion.

v2 = 2*m1*u1/(m1+m2)

= 2*0.3*7.16/(0.3+1.3)

= 2.69 m/s <<<<<<<---------Answer

c) v1 = (m1-m2)*u1/(m1+m2)

= (0.3 - 1.3)*7.16/(0.3+1.3)

= -4.48 m/s <<<<<<<---------Answer

d) Again apply energy coservation

0.5*k*x^2 = 0.5*m*v^2

x = v*sqrt(m/k)

= 4.48*sqrt(0.3/1900)

= 0.056 m or 5.6 cm <<<<<<<---------Answer

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