Two particles A and B, of mass 5m kg and 2m kg respectively, are moving in opposite directions along the same straight horizontal line. The particles collide directly. Immediately before the collision, the speeds of A and B are 3 m sâˆ’1 and 4 m sâˆ’1 respectively. The direction of motion of A is unchanged by the collision. Immediately after the collision, the speed of A is 0.8 m sâˆ’1. (a) Find the speed of B immediately after the collision. In the collision, the magnitude of the impulse exerted on A by B is 3.3 N s. (b) Find the value of m.

A-Level Edexcel Maths Mechanics Newtons Second Law: Two particles A and B, of mass 5

Two particles A and B, of mass 5m kg and 2m kg respectively, are moving in opposite directions along the same straight horizontal line - Edexcel - A-Level Maths Mechanics - Question 1 - 2012 - Paper 1

Question 1

Two-particles-A-and-B,-of-mass-5m-kg-and-2m-kg-respectively,-are-moving-in-opposite-directions-along-the-same-straight-horizontal-line-Edexcel-A-Level Maths Mechanics-Question 1-2012-Paper 1.png

Two particles A and B, of mass 5m kg and 2m kg respectively, are moving in opposite directions along the same straight horizontal line. The particles collide directl... show full transcript

Worked Solution & Example Answer:Two particles A and B, of mass 5m kg and 2m kg respectively, are moving in opposite directions along the same straight horizontal line - Edexcel - A-Level Maths Mechanics - Question 1 - 2012 - Paper 1

Step 1

Find the speed of B immediately after the collision.

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Answer

To find the speed of B immediately after the collision, we apply the principle of conservation of linear momentum. The formula for momentum before collision is:

$ext{Total Initial Momentum} = ext{Total Final Momentum}$

Before the collision:

Momentum of A = mass × velocity = $5m imes 3 = 15m$ kg m/s (moving in one direction)
Momentum of B = mass × velocity = $2m imes -4 = -8m$ kg m/s (moving in the opposite direction)

Thus, the total initial momentum is: $15m - 8m = 7m$

After the collision:

Momentum of A = $5m imes 0.8 = 4m$ kg m/s
Let the speed of B after the collision be v:
Momentum of B = $2m imes v = 2mv$ kg m/s

Thus, the total final momentum is: $4m + 2mv$

Setting total initial momentum equal to total final momentum gives: $7m = 4m + 2mv$ Solving for v, we have: $3m = 2mv$ $v = rac{3m}{2m} = 1.5 ext{ m/s}$

Thus, the speed of B immediately after the collision is 1.5 m/s.

Step 2

Find the value of m.

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Answer

To find m, we can use the impulse-momentum theorem. The impulse exerted on A by B is given as 3.3 N s. The impulse is defined as the change in momentum:

$ext{Impulse} = ext{Change in Momentum}$

For particle A:

Initial momentum of A = $5m imes 3 = 15m$ kg m/s
Final momentum of A = $5m imes 0.8 = 4m$ kg m/s

Therefore, the change in momentum for A is: $ext{Change in Momentum} = 4m - 15m = -11m$

Setting this equal to the impulse: $-11m = -3.3$ Thus: $m = rac{3.3}{11} = 0.3$

Therefore, the value of m is 0.3.

Two particles A and B, of mass 5m kg and 2m kg respectively, are moving in opposite directions along the same straight horizontal line - Edexcel - A-Level Maths Mechanics - Question 1 - 2012 - Paper 1

Worked Solution & Example Answer:Two particles A and B, of mass 5m kg and 2m kg respectively, are moving in opposite directions along the same straight horizontal line - Edexcel - A-Level Maths Mechanics - Question 1 - 2012 - Paper 1

Find the speed of B immediately after the collision.

Find the value of m.

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