Question

In the figure below, block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2.30 m and then collides with stationary block 2, which has mass m2= 2.00m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction ?k is 0.600 and comes to a stop in distance d within thatregion.

(a) What is the value of distance d if the collision is elastic?
m

(b) What is the value of distance d if the collision is completely inelastic?
m

Answer 1

KE = 1/2(m1)v^2 = (m1)gh
v = sqrt(2gh) = sqrt(2 * 9.8 * 2.5) = 7 m/s

Thus, the kinetic energy is 24.5 * m1 J and the momentum is 7 * m1 kg*m/s.

(a) In an elastic collision, momentum and KE are conserved. So you have 2 equations for 2 unknowns (since this is a 1D collision) - masses can cancel outsince m2 = 2 * m1
24.5 * m1 = 1/2(m1)(v1)^2 + 1/2(m2)(v2)^2
24.5 = 1/2 * v2^2 + v2^2
49 = (v1)^2 + 2 * v2^2

and

7 * m1 = (v1)(m1) + (v2)(m2)
7 = v1 + 2 * v2

Substitute to solve for v1 and v2:
49 = (v1)^2 + 2 * v2^2
49 = (7 - 2 * v2)^2 + 2 * v2^2
49 = 49 - 28 * v2 + 6 * v2^2
0 = 6 * v2^2 - 28 * v2 = 3 * v2^2 - 14 * v2
0 = v2(3 * v2 - 14)

v2 = 0 OR v2 = 14/3

0 makes no sense, so we get 14/3 m/s. Now, it hits the region with friction - we need the acceleration due to the frictional force and solve for d usingkinematics

F = 0.6 * (m2 * g)
a = F / m2 = 0.6 * g = 5.88

v = v0 + at
0 = 14/3 m/s - 5.88t
t = 0.79

x = x0 + v0t + (1/2)at^2
x = 5.55 m = d


(b) Almost the same thing, but KE is not conserved.

7 * m1 = v2 * (m1 + m2) = v2 * (3 * m1)
7 = 3 * v2
v2 = 7/3

Now, use the same formula for calculating the distance as before but just change the initial velocity.

Answer 2

a.0.3806 m
b.0.8518 m

Answer 3

you get is:

d = 3.7 m