Particle P has mass 3 kg and particle Q has mass 2 kg. The particles are moving in opposite directions on a smooth horizontal plane when they collide directly. Immediately before the collision, P has speed 3 m s⁻¹ and Q has speed 2 m s⁻¹. Immediately after the collision, both particles move in the same direction and the difference in their speeds is 1 m s⁻¹. (a) Find the speed of each particle after the collision. (b) Find the magnitude of the impulse exerted on P by Q.

A-Level Edexcel Maths Mechanics Further Forces & Newtons Laws: Particle P has mass 3 kg and par

Particle P has mass 3 kg and particle Q has mass 2 kg - Edexcel - A-Level Maths Mechanics - Question 2 - 2011 - Paper 1

Question 2

Particle P has mass 3 kg and particle Q has mass 2 kg. The particles are moving in opposite directions on a smooth horizontal plane when they collide directly. Immed... show full transcript

Worked Solution & Example Answer:Particle P has mass 3 kg and particle Q has mass 2 kg - Edexcel - A-Level Maths Mechanics - Question 2 - 2011 - Paper 1

Step 1

Find the speed of each particle after the collision.

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Answer

To solve for the speeds of particles P and Q after the collision, we apply the principle of conservation of linear momentum:

$3 \times 3 + (-2) \times 2 = 3v + 2(v+1)$

Where:

The left side of the equation represents the momentum before the collision.
The right side represents the momentum after the collision, where v is the speed of particle P and (v + 1) is the speed of particle Q after the collision.

Now simplifying this equation:

$9 - 4 = 3v + 2v + 2$

$5 = 5v + 2$

Subtracting 2 from both sides:

$3 = 5v$

Now, solving for v:

$v = \frac{3}{5} = 0.6 \, \text{m s}^{-1}$

Thus, the speed of particle P after the collision is 0.6 m s⁻¹.

For particle Q:

Since the difference in speeds after the collision is 1 m s⁻¹, we have:

$v+1 - v = 1$

Thus, the speed of particle Q after the collision is:

$v_Q = v + 1 = 0.6 + 1 = 1.6 \, \text{m s}^{-1}$

In conclusion:

The speed of particle P is 0.6 m s⁻¹.
The speed of particle Q is 1.6 m s⁻¹.

Step 2

Find the magnitude of the impulse exerted on P by Q.

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Answer

The impulse exerted on particle P by particle Q can be calculated using the formula:

$\text{Impulse} = \Delta p = m(v_f - v_i)$

Where:

m is the mass of particle Q (2 kg).
$v_f$ is the final speed of P (0.6 m s⁻¹).
$v_i$ is the initial speed of P (3 m s⁻¹).

Now substituting the values:

$\Delta p = 2(0.6 - 3) = 2(-2.4) = -4.8 \, \text{Ns}$

The negative sign indicates that the impulse direction is opposite to the initial direction of P. The magnitude of the impulse is:

$|\Delta p| = 4.8 \, \text{Ns}$

Thus, the magnitude of the impulse exerted on P by Q is 4.8 Ns.

Particle P has mass 3 kg and particle Q has mass 2 kg - Edexcel - A-Level Maths Mechanics - Question 2 - 2011 - Paper 1

Worked Solution & Example Answer:Particle P has mass 3 kg and particle Q has mass 2 kg - Edexcel - A-Level Maths Mechanics - Question 2 - 2011 - Paper 1

Find the speed of each particle after the collision.

Find the magnitude of the impulse exerted on P by Q.

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