Question

A block of mass m= 2.50 kg slides down a 30.0° incline which is 3.60 m high. At the bottom, it strikes a block of mass m = 7.00 kg which is at rest on a horizontal surface. (Assume a smooth transition at the bottom of the incline.) If the collision is elastic, and friction can be ignored, determine (a) the speeds of the two blocks after the collision, and (b) how far back up the incline the smaller mass will go.

Answer 1

Answer:

(a) The speed of the lighter block is $v_{2x} = 3.45~m/s.$

   The speed of the heavier block is  $v_{3x} = 3.83~m/s.$

(b) The smaller block goes up to 0.61 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.

$K_1 + U_1 = K_2 + U_2\\0 + mgh = \frac{1}{2}mv_1^2 + 0\\(2.5)(9.8)(3.6) = \frac{1}{2}(2.5)v_1^2\\v_1 = 8.4 ~m/s$

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.

$P_1 = P_2\\mv_{1x} = mv_{2x} + Mv_{3x}\\(2.5)(8.4\cos(30^\circ)) = (2.5)v_{2x} + (7)v_{3x}\\18.19 = 2.5v_{2x} + 7v_{3x}$

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

$K_{initial} = K_{final}\\\frac{1}{2}mv_1^2 = \frac{1}{2}mv_{2}^2 + \frac{1}{2}Mv_{3x}^2\\(2)(8.4)^2 = (2)v_{2}^2 + (7)v_{3x}^2\\141.12 = 2v_{2}^2 + 7v_{3x}^2$

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:

$v_{3x} = \frac{2m}{(m+M)}v_{1x} = \frac{2(2.5)}{(2.5 + 7)}(8.4\cos(30^\circ)) = 3.83~m/s$

Therefore from the first equation,

$18.19 = 2.5v_{2x} + 7v_{3x}$

$18.19 = 2.5v_{2x} + 7(3.83)\\\\v_{2x}= -3.45m/s$

Part C:

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

$K_1 + U_1 = K_2 + U_2\\\frac{1}{2}mv_{2x}^2 + 0 = 0 + mgH\\\frac{1}{2}(2.5)(-3.45)^2 = (2.5)(9.8)H\\H = 0.61 ~m$