A stone S is sliding on ice - Edexcel - A-Level Maths Mechanics - Question 6 - 2005 - Paper 1

To find the speed of S at C, we again use the formula: $v^2 = u^2 + 2as$ Where:

• $u = 20 \, \text{m/s}$ (initial speed at A)
• $s = 30 \, \text{m}$ (distance to C)
• $a = -3 \, \text{m/s}^2$ (deceleration)

Substituting the values: $v^2 = 20^2 + 2(-3)(30)$

This simplifies to: $v^2 = 400 - 180$ $v^2 = 220$

Taking the square root gives: $v = \sqrt{220} \, \text{m/s} \approx 14.8 \, \text{m/s}$

Thus, the speed of S at C is approximately 14.8 m/s.

Using the formula for force due to mass and acceleration: $F = ma$ Where:

• $F = 0.3 \, \text{N}$ (constant resistance)
• $m$ is the mass, and
• $a = 3 \, \text{m/s}^2$ (deceleration)

Rearranging yields: $m = \frac{F}{a} = \frac{0.3}{3} = 0.1 \, \text{kg}$

Thus, the mass of S is confirmed to be 0.1 kg.

First, we calculate the net force acting on S after it rebounds:

• Impulse from the wall: 2.4 N s
• Resistance: 0.3 N

The total force acting after rebounding: $F_{net} = w - R = 2.4 - 0.3 = 2.1 \, \text{N}$

Using $w$, we can find the acceleration: $w = ma$

• Let $m = 0.1 \, \text{kg}$ (mass)
• Rearranging, we get: $a = \frac{w}{m} = \frac{2.1}{0.1} = 21 \, \text{m/s}^2$

Next, we apply the kinematic equation to find the time, where the initial velocity ($u$) after rebounding is approximately 14.8 m/s:

\text{where, } g = 21 \, \text{m/s}^2$$ Thus, $$0 = 14.8 - 21t$$ Solving for $t$: $$21t = 14.8 \ t \approx \frac{14.8}{21} \approx 0.706 \text{seconds},$$ Finally, total time considering both directions gives: $$t_{total} = 0.706 + \frac{14.8}{3} \to t \approx 3.06 \, \text{seconds}$$ So, the total time between the instant that S rebounds from the wall and comes to rest is approximately 3.06 seconds.