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 Forces: 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.

96%

114 rated

Answer

To find the speed of each particle after the collision, we can apply the principle of conservation of linear momentum. We represent the speed of particle P after the collision as $v$ and the speed of particle Q after the collision as $v + 1$ (since the difference in their speeds after the collision is 1 m/s).

The total momentum before the collision must equal the total momentum after the collision:

$3 \text{ kg} \times 3 \text{ m/s} + 2 \text{ kg} \times (-2 \text{ m/s}) = 3 \text{ kg} \times v + 2 \text{ kg} \times (v + 1)$

Calculating the left side: $9 - 4 = 5 \text{ kg m/s}$

Setting up the equation: $5 = 3v + 2(v + 1)$

This simplifies to: $5 = 3v + 2v + 2$ $5 = 5v + 2$ $3 = 5v$ $v = 0.6 \text{ m/s}$

Thus, the speed of particle P after the collision is $0.6 \text{ m/s}$ and the speed of particle Q is: $v + 1 = 1.6 \text{ m/s}$ .

Step 2

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

99%

104 rated

Answer

The impulse exerted on particle P by particle Q can be determined using the change in momentum. Impulse is defined as:

$ext{Impulse} = ext{Change in Momentum} = m(v_{final} - v_{initial})$

For particle P:

Mass $m = 3 \text{ kg}$
Initial speed $v_{initial} = 3 \text{ m/s}$
Final speed $v_{final} = 0.6 \text{ m/s}$

Thus, the change in momentum for particle P is: $ext{Impulse} = 3(0.6 - 3)$ $= 3(-2.4)$ $= -7.2 \text{ Ns}$

Since impulse is a magnitude, we take the absolute value: $| ext{Impulse}| = 7.2 \text{ Ns}$ .

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.

Join the A-Level students using SimpleStudy...