Question

A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0

∘
incline which is 3.60 m high. At the bottom, it strikes a block of mass M=7.00M=7.00 kg which is at rest on a horizontal surface, (Assuming a smooth transition at the bottom of the incline.) If the collision is elastic, and friction can be ignored, determine the speeds of the two blocks (v_mv
m
​
and v_Mv
M
​
) after the collision.

Answer 1

`v_m \approx -4.38\; \rm m \cdot s^(-1)` (moving toward the incline.)

`v_M \approx 4.02\; \rm m \cdot s^(-1)` (moving away from the incline.)

(Assumption:
`g = 9.81\; \rm m \cdot s^(-2)`.)

Step-by-step explanation:

If
`g = 9.81\; \rm m \cdot s^(-2)`, the potential energy of the block of
`m = 2.20\; \rm kg` would be
`m \cdot g\cdot h = 2.20\; \rm kg * 9.81\; \rm m \cdot s^(-2) * 3.60\; \rm m \approx 77.695\; \rm J` when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this
`m = 2.20\; \rm kg\!` block right before the collision would also be approximately
`77.695\; \rm J`.

Calculate the velocity of that
`m = 2.20\; \rm kg` based on its kinetic energy:

`\displaystyle v_m(\text{initial}) = \sqrt{\frac{2* (\text{Kinetic Energy})}{m}} \approx \sqrt{(2 * 77.695\; \rm J)/(2.20\; \rm kg)} \approx 8.4043\; \rm m \cdot s^(-1)}`.

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

`p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg * 8.4043\; \rm m \cdot s^(-1) \approx 18.489\; \rm kg \cdot m \cdot s^(-1)`.

`p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg * 0\; \rm m \cdot s^(-1) \approx 0\; \rm kg \cdot m \cdot s^(-1)`.

Sum of the momentum of each block right before the collision: approximately
`18.489\; \rm kg \cdot m \cdot s^(-1)`.

Sum of the momentum of each block right after the collision:
`(m\cdot v_m + m \cdot v_M)`.

For momentum to conserve in this collision,
`v_m` and
`v_M` should ensure that
`m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^(-1)`.

Kinetic energy of the two blocks right before the collision: approximately
`77.695\; \rm J` and
`0\; \rm J`. Sum of these two values: approximately
`77.695\; \rm J\!`.

Sum of the energy of each block right after the collision:

`\displaystyle \left((1)/(2)\, m \cdot {v_m}^2 + (1)/(2)\, M \cdot {v_M}^2\right)`.

Similarly, for kinetic energy to conserve in this collision,
`v_m` and
`v_M` should ensure that
`\displaystyle (1)/(2)\, m \cdot {v_m}^2 + (1)/(2)\, M \cdot {v_M}^2 \approx 77.695\; \rm J`.

Combine to obtain two equations about
`v_m` and
`v_M` (given that
`m = 2.20\; \rm kg` whereas
`M = 7.00\; \rm kg`.)

`\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^(-1) \\ & (1)/(2)\, m \cdot {v_m}^2 + (1)/(2)\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right.`.

Solve for
`v_m` and
`v_M` (ignore the root where
`v_M = 0`.)

`\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^(-1) \\ & v_M \approx 4.02\; \rm m \cdot s^(-1)\end{aligned}\right.`.

The collision flipped the sign of the velocity of the
`m = 2.20\; \rm kg` block. In other words, this block is moving backwards towards the incline after the collision.