A small block of mass M is released from rest at the top of the curved frictionl

Question

A small block of mass M is released from rest at the top of the curved frictionless ramp shown

above. The block slides down the ramp and is moving with speed 3.5v0 when it collides with a

larger block of mass 1.5M at rest at the bottom of the incline. The larger block moves to the right

at a speed of 2v0 immediately after the collision. Express your answers to the following questions

in terms of the given quantities and fundamental constants.

(a) Determine the height h of the ramp from which the small block was released.

(b) Determine the speed of the small block after the collision.

(c) The larger block slides a distance D before coming to rest. Determine the value of the

coefficient of kinetic friction ?? between the larger block and the surface on which it slides.

(d) Indicate whether the collision between the two blocks is elastic or inelastic. Justify your

answer.

Can you please just supply formulas as if there is no numbers involved, thank you for your help

A small block of mass M is released from rest at the top of the curved frictionless ramp shown above. The block slides down the ramp and is moving with speed 3.5v0 when it collides with a larger block of mass 1.5M at rest at the bottom of the incline. The larger block moves to the right at a speed of 2v0 immediately after the collision. Express your answers to the following questions in terms of the given quantities and fundamental constants. Determine the height h of the ramp from which the small block was released. Determine the speed of the small block after the collision. The larger block slides a distance D before coming to rest. Determine the value of the coefficient of kinetic friction ?? between the larger block and the surface on which it slides. Indicate whether the collision between the two blocks is elastic or inelastic. Justify your answer.

Explanation / Answer

m1 = M, u1 = 3.5*vo
m2 = 1.5*M, u2 = 0, v2 = 2*vo

a) 3.5*vo = sqrt(2*g*h)

h = 3.5^2*vo^2/2*g

h = 0.625*vo

b)

let vo is the speed of m1 after the collision

linear momentum is conserved

m1*u1 + m2*u2 = m1*v1 + m2*v2

M*3.5*vo + 0 = M*v1 + 1.5*M*2*vo

M*v1 = M*3.5*vo - 3*M*vo


v1 = 0.5*vo

c)

v2'^2 - v2^2 = 2*a*s

0- (2*Vo)^2 = -2*mue*M*g*D

==> mue = 2*vo^2/(M*g*D)

d)the collision is inelastic. Because, there is loss of kinetic energy in the collision.

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