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

To find the speed at C, we again use the formula:

$v^2 = u^2 + 2as$

where:

• $u = 16 \, \text{m/s}$ (speed at B)
• $a = -3 \, \text{m/s}^2$ (deceleration)
• $s = 30 \, \text{m}$ (distance BC)

Now, we rearrange:

$v^2 = 16^2 - 2 \cdot 3 \cdot 30$

Calculating:

$v^2 = 256 - 180 = 76$

So, $v = \sqrt{76} \approx 8.72 \, \text{m/s}$

Thus, the speed of S at C is approximately 8.72 m/s.

Using the relation between force, mass, and acceleration:

$F = ma$

Given that the resistance force $F = 0.3 \, N$ and we found the deceleration $a = 3 \, m/s^2$, we can solve for the mass $m$:

$0.3 = m \cdot 3 \Rightarrow m = \frac{0.3}{3} = 0.1 \, kg$

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

After rebounding, the stone moves against the constant resistance. The impulse given by the wall is 2.4 N s, which affects its motion.

The resulting force will be: $w = 0.1 \cdot (0.3 + 20) = 2.4 \, N$

Using Newton's second law for the resistance, we find the acceleration:

$a = \frac{w}{m} = \frac{2.4}{0.1} = 24 \, m/s^2$

Next, using the equation of motion:

$v = u + at$

Where $u = 20 \, m/s$ (speed during rebound), the final speed $v = 0$, and $a = -24 \, m/s^2$:

Setting these values we get:

$0 = 20 - 24t \Rightarrow 24t = 20 \Rightarrow t = \frac{20}{24} = 0.83 \, s$

Thus, the time between the instant that S rebounds from the wall and the instant that S comes to rest is approximately 0.83 seconds.