HELP PLZ Block 1, mass m, slides from rest down the inside of a bowl shaped cont

Question

HELP PLZ

Block 1, mass m, slides from rest down the inside of a bowl shaped container, starting from a vertical height h. The surface of the container is frictionless. At the bottom, it collides inelastically with Block 2, mass 1/3 m, and the two stick together. To what vertical height does the combined mass slide? Answer in terms of h. Show work & answer here: A ball with v0 = 15 m/s collides with two identical balls which are lined up vertically as in the diagram below. The incoming ball is aimed directly at the contact point between the two balls. The collision is elastic, and there are no external forces acting. Find the velocities of all three balls after the collision. Note: The directions of the balls after the collision can be found by considering the direction of the impulse each receives at the moment of impact. After collision: V1 = V2 = V3 = Show work here.

Explanation / Answer

Assume an initial velocity of u for the block of mass m just before the inelastic collision.

Since there has been no disipation of energy till this point,

(.5)*m*(u^2)=m*g*h ---- (1)

Now, though the collision is inelastic, momentum is conserved. Hence, for the final velocity after collision,

m*u= (m+(m/3))*v, i.e., v= (3/4)*u -----(2)

From 1 & 2, we have final height h' as

4/3 *m*g*h' = .5 *(4m/3) * v^2 =(9/16)(.5*m*u^2)*(4/3)= (36/48)*m*g*h

h' = (9/16) *h

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

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