A 0.3 kg block compresses a spring of spring constant 1000 N/m by 0.13 m. After

Question

A 0.3 kg block compresses a spring of spring constant 1000 N/m by 0.13 m. After being released from rest, the block slides along a smooth, horizontal and frictionless surface before colliding elastically with a 1.5 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 0.3 kg block just before striking the 1.5 kg block? V=7.51m/s Correct
B.What is the velocity of the 1.5 kg block after the collision?
C.What is the velocity of the 0.3 kg block after the collision?
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

spring energy = 1/2kx^2

k.e = 1/2mv^2

now, 1/2kx^2 = 1/2mv^2

v = 7.51 m/s

by conservation of momentum

0.3*7.51 = 0.3V1 + 1.5V2

as elastic collision

1 = 7.51/(v2-V1)

V2-V1 = 7.51

by solving

B.   V2 = 2.50 m/s

C. V1 = -5.01 m/s

maximum compression = 8.67 cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.