COM CYU 13

Physics Solution – Q13

Question 13: Block on Accelerating Cart

1. System Analysis & Diagram

We consider the motion of the block $A$ relative to the cart $B$. The cart is placed on a frictionless floor, but there is friction between the block and the cart. When the block lands on the cart, it has a velocity relative to the cart. It travels forward, hits the elastic obstruction, and returns to the rear end.

B (M) A m f = μmg f’ = μmg l = 1.0 m

2. Relative Acceleration

Since we need to find the time spent on the cart, it is most convenient to work in the non-inertial frame of the cart $B$.

Forces acting horizontally:
Friction force $f = \mu mg$. This force acts backward on block $A$ and forward on cart $B$.

Accelerations relative to the ground:

  • Acceleration of block $A$: $a_A = -\frac{f}{m} = -\mu g$ (retardation)
  • Acceleration of cart $B$: $a_B = \frac{f}{M} = \frac{\mu mg}{M}$

Relative Acceleration ($a_{rel}$):
The acceleration of the block relative to the cart is: $$ a_{rel} = a_A – a_B = -\mu g – \frac{\mu mg}{M} = -\mu g \left( 1 + \frac{m}{M} \right) $$ $$ a_{rel} = -\mu g \left( \frac{M + m}{M} \right) $$ This relative acceleration acts opposite to the direction of motion of the block relative to the cart. Since the collision is perfectly elastic, the speed relative to the cart is preserved during the bounce, but the direction reverses. The friction continues to oppose the relative motion throughout. Thus, the magnitude of retardation remains constant.

3. Calculation of Time

The block starts at the left end, travels distance $l$ to the obstruction, bounces, and travels distance $l$ back to the rear end.

  • Total relative distance covered, $s_{rel} = 2l$.
  • Final relative velocity, $v_{rel} = 0$ (since it stops at the rear end).
  • We need to find time $t$.

Using the kinematic equation of motion reversed (considering motion from stop backwards to start): $$ s = \frac{1}{2} |a_{rel}| t^2 $$ $$ 2l = \frac{1}{2} \left[ \mu g \left( \frac{M + m}{M} \right) \right] t^2 $$

Solving for $t$: $$ t^2 = \frac{4l \cdot M}{\mu g (M + m)} $$ $$ t = 2 \sqrt{\frac{Ml}{(M + m)\mu g}} $$

Substitute values:
$M = 8.0 \, \text{kg}$, $m = 2.0 \, \text{kg}$, $l = 1.0 \, \text{m}$, $\mu = 0.5$, $g = 10 \, \text{m/s}^2$. $$ t = 2 \sqrt{\frac{8.0 \times 1.0}{(8.0 + 2.0) \times 0.5 \times 10}} $$ $$ t = 2 \sqrt{\frac{8}{10 \times 5}} = 2 \sqrt{\frac{8}{50}} = 2 \sqrt{\frac{4}{25}} = 2 \times \frac{2}{5} = \frac{4}{5} = 0.8 \text{ s} $$

4. Loss of Mechanical Energy

The loss of mechanical energy is equal to the work done by friction against the relative sliding motion. $$ W_{friction} = f \times s_{rel} $$ $$ \Delta E = (\mu mg) \times (2l) = 2 \mu m g l $$

Substitute values: $$ \Delta E = 2 \times 0.5 \times 2.0 \times 10 \times 1.0 $$ $$ \Delta E = 20 \, \text{J} $$

Final Answer

Time spent: $$ t = 2\sqrt{\frac{Ml}{(m+M)\mu g}} $$ (Substituting values yields 0.8 s).

Energy Loss: $$ 2\mu mgl = 20 \, \text{J} $$

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