Question

A block of mass m = 2.20 kg slides down a 30.0° incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 6.80 kg which is at rest on a horizontal surface, (Assume an elastic collision, and ignore friction.)
(a) Determine the speeds of the two blocks after the collision.
lighter block
m/s
heavier block
m/s

(b) Determine how far back up the incline the smaller mass will go.
m

Answer 1

Answer:

(a) The speed of the lighter block is $v_{2x} = 3.7~m/s$.

    The speed of the heavier block is $v_{3x} = 3.55~m/s$.

(b) The smaller block goes up to 0.69 m.

Explanation:

We will divide this question into three parts: Part A is for the smaller mass from the top of the incline to the collision. Part B is the collision. And Part C is for the smaller mass from the bottom to the highest point it can achieve.

In order to solve this question, I will assume that the smaller mass is initially at rest.

Part A:

We will use conservation of energy.

$K_1 + U_1 = K_2 + U_2\\0 + mgh = \frac{1}{2}mv_1^2 + 0\\(2.2)(9.8)(3.6) = \frac{1}{2}(2.2)v_1^2\\v_1 = 8.4 ~m/s$

This is the speed of the smaller mass just before the collision. The velocity of the mass is directed 30° above horizontal, since the mass is sliding down the incline.

Part B:

Momentum is a vector identity, so the x- and y-components of momentum are to be investigated separately. Since the collision occurs at the horizontal surface, only the x-component of momentum is conserved.

$P_1 = P_2\\mv_{1x} = mv_{2x} + Mv_{3x}\\(2.2)(8.4\cos(30^\circ)) = (2.2)v_{2x} + (6.8)v_{3x}\\16 = 2.2v_{2x} + 6.8v_{3x}$

During the collision kinetic energy is also conserved. Since kinetic energy is a scalar quantity, we don't have to separate its components.

$K_{initial} = K_{final}\\\frac{1}{2}mv_1^2 = \frac{1}{2}mv_{2}^2 + \frac{1}{2}Mv_{3x}^2\\(2.2)(8.4)^2 = (2.2)v_{2}^2 + (6.8)v_{3x}^2\\155.23 = 2.2v_{2}^2 + 6.8v_{3x}^2$

The following relation will be used when combining the two equations:

$v_{2x} = v_2\cos{30^\circ}$

The following equation is useful for combining the two equations:

$v_{3x} = \frac{2m}{(m+M)}v_{1x} = \frac{2(2.2)}{(2.2 + 6.8)}(8.4\cos(30^\circ)) = 3.55~m/s$

Therefore from the first equation,

$16 = 2.2v_{2x} + 6.8v_{3x} = 2.2v_{2x} + 6.8(3.55) \\v_{2x} = -3.7~m/s$

Part C:

We will again use the conservation of energy to find the highest point that the mass can go:

$K_1 + U_1 = K_2 + U_2\\\frac{1}{2}mv_{2x}^2 + 0 = 0 + mgH\\\frac{1}{2}(2.2)(-3.7)^2 = (2.2)(9.8)H\\H = 0.69 ~m$