A 4.0 kg block moving to the right with a speed of 10 m/s collides with a 5.0 kg

Question

A 4.0 kg block moving to the right with a speed of 10 m/s collides with a 5.0 kg block moving to the right with a speed of 4.0 m/s. The duration of the collision is 10.0 milliseconds. After the collision, the 5.0 kg block moves to the right with a speed of 8.0 m/s. Find: the final speed of the 4.0 kg block, the coefficient of restitution for the collision, the impulse delivered to the 5.00 kg mass during the collision, the average force exerted on the 5.00 kg mass during the collision, and, the energy lost (if any) in the collision.

Explanation / Answer

Use the law of conservation of momentum to find the velocity of 4kg block.

V= m1u1+m2u2-m2v2/m1

= 5 m/s

The value of e is,

e = 8-4/4-10 = -2/3=-0.666

The change in momentum is equal to impulse

Impulse= m2(v2-u2)=20 N.s

The average force on 5 kg is

F=dp/dt= 20/10*10^-3 = 2000 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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