Particle P has mass m kg and particle Q has mass 3m kg. The particles are moving in opposite directions along a smooth horizontal plane when they collide directly. Immediately before the collision P has speed 4u m s⁻¹ and Q has speed ku m s⁻¹, where k is a constant. As a result of the collision the direction of motion of each particle is reversed and the speed of each particle is halved. (a) Find the value of k. (b) Find, in terms of m and u, the magnitude of the impulse exerted on P by Q.

A-Level Edexcel Maths Mechanics Forces: Particle P has mass m kg and par

Particle P has mass m kg and particle Q has mass 3m kg - Edexcel - A-Level Maths Mechanics - Question 2 - 2010 - Paper 1

Question 2

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

Worked Solution & Example Answer:Particle P has mass m kg and particle Q has mass 3m kg - Edexcel - A-Level Maths Mechanics - Question 2 - 2010 - Paper 1

Step 1

Find the value of k.

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Answer

To solve for k, we apply the principle of conservation of momentum before and after the collision. The total momentum before the collision is:

$p_{initial} = m(4u) + 3m(-ku)$

The total momentum after the collision is:

$p_{final} = m(-2u) + 3m(\frac{-ku}{2})$

Setting the two equal, we have:

$m(4u) - 3mku = -2mu - rac{3mku}{2}$

Canceling m and reorganizing:

$4u - 3ku = -2u - \frac{3ku}{2}$

Bringing terms involving k on one side gives:

$4u + 2u = 3ku - \frac{3ku}{2} \implies 6u = \frac{3ku}{2}$

Multiplying through by 2 to eliminate the fraction:

$12u = 3ku \implies k = \frac{4}{3}$

Step 2

Find, in terms of m and u, 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 defined as the change in momentum of particle P. The initial momentum of P before the collision is:

$p_{P, initial} = mu(4)$

The final momentum of P after the collision, using k from part (a), is:

$p_{P, final} = m(-2)$

The change in momentum, which is the impulse (I), is given by:

$I = p_{P, final} - p_{P, initial}$

Substituting the expressions we found:

$I = m(-2) - m(4) = -6mu$

Thus, the magnitude of the impulse is:

$|I| = 6mu$

Particle P has mass m kg and particle Q has mass 3m kg - Edexcel - A-Level Maths Mechanics - Question 2 - 2010 - Paper 1

Worked Solution & Example Answer:Particle P has mass m kg and particle Q has mass 3m kg - Edexcel - A-Level Maths Mechanics - Question 2 - 2010 - Paper 1

Find the value of k.

Find, in terms of m and u, the magnitude of the impulse exerted on P by Q.

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