A body of mass m_1 = 15 kg slides down a smooth incline of…

A body of mass $m_1 = 15\ \mathrm{kg}$ slides down a smooth incline of angle $\alpha = 60^\circ$ from a height $h = 10\ \mathrm{m}$ and at the bottom plunges into a cart full of sand of mass $m_2 = 90\ \mathrm{kg}$, standing on a smooth horizontal floor. With what speed do they move on together?

Explanation of Method

A body sliding into a sand cart.

Step 1: Speed at the bottom

\[ v_1 = \sqrt{2gh} \approx 14.0\ \mathrm{m/s} \ \text{(along the incline)} \]

Step 2: Horizontal momentum

During the impact the ground pushes up on the cart, so only the horizontal momentum is conserved: $m_1 v_1\cos\alpha = (m_1 + m_2)\,v$.

\[ v = \frac{15 \cdot 7.0}{105}. \]

\[v \approx 1.0\ \mathrm{m/s}\]

$h = 10\ \mathrm{m}$ $90\ \mathrm{kg}$ $v = 1\ \mathrm{m/s}$ $v_1 = 14\ \mathrm{m/s}$ $v_1\cos60^\circ = 7$ $60^\circ$

Only the horizontal part of the body's velocity (7 m/s) is shared with the cart; the vertical part is stopped by the ground through the cart's wheels.

Result

\[\boxed{\,v \approx 1.0\ \mathrm{m/s}\,}\]

Remark

Check. The original result is correct. Of the body's 1471 J of kinetic energy only 52 J remain as motion of the cart; the rest goes into the sand and the ground.