Two particles, P and Q, have masses 2m and 3m respectively. They are moving towards each other in opposite directions on a smooth horizontal plane when they collide directly. Immediately before they collide the speed of P is 4u and the speed of Q is 3u. As a result of the collision, Q has its direction of motion reversed and is moving with speed u. (a) Find the speed of P immediately after the collision. (b) State whether or not the direction of motion of P has been reversed by the collision. (c) Find the magnitude of the impulse exerted on P by Q in the collision.

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Two particles, P and Q, have masses 2m and 3m respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2017 - Paper 1

Question 2

Two particles, P and Q, have masses 2m and 3m respectively. They are moving towards each other in opposite directions on a smooth horizontal plane when they collide ... show full transcript

Worked Solution & Example Answer:Two particles, P and Q, have masses 2m and 3m respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2017 - Paper 1

Step 1

Find the speed of P immediately after the collision.

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Answer

To find the speed of P immediately after the collision, we will apply the principle of conservation of linear momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

Before the Collision:
- Momentum of P:
  $p_P = (2m)(4u) = 8mu$
- Momentum of Q: $p_Q = (3m)(-3u) = -9mu$
- Total momentum before collision: $p_{initial} = 8mu - 9mu = -mu$
After the Collision:
- Let the speed of P after the collision be v.
- Momentum of P after the collision: $p'_P = (2m)(v) = 2mv$
- Since Q reverses direction, its momentum after collision: $p'_Q = (3m)(-u) = -3mu$
- Total momentum after collision: $p_{final} = 2mv - 3mu$
Setting Up the Equation: Using conservation of linear momentum: $p_{initial} = p_{final}$ $-mu = 2mv - 3mu$
Solving for v: Rearranging gives us: $2mv = -mu + 3mu$ $2mv = 2mu$ $v = u$

Thus, the speed of P immediately after the collision is u.

Step 2

State whether or not the direction of motion of P has been reversed by the collision.

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Answer

From our previous calculations, we can see that the speed of P after the collision is u, which is in the same direction as its original motion. Therefore, the direction of motion of P has not been reversed by the collision.

Step 3

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

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Answer

Impulse is defined as the change in momentum. We can calculate the impulse exerted on P by Q as follows:

Change in Momentum for P:
- Initial momentum of P: $p_{initial} = 8mu$
- Final momentum of P after the collision: $p_{final} = 2mv$ From the previous step, we found that v = u, therefore:
- Final momentum: $p_{final} = 2m(u) = 2mu$
Calculating the Impulse: Impulse (J) is given as: $J = p_{final} - p_{initial}$ Substituting we get: $J = 2mu - 8mu = -6mu$
- The magnitude of impulse is: $|J| = 6mu$

Thus, the magnitude of the impulse exerted on P by Q in the collision is 6mu.

Two particles, P and Q, have masses 2m and 3m respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2017 - Paper 1

Worked Solution & Example Answer:Two particles, P and Q, have masses 2m and 3m respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2017 - Paper 1

Find the speed of P immediately after the collision.

State whether or not the direction of motion of P has been reversed by the collision.

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

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