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Two particles A and B are moving on a smooth horizontal plane - Edexcel - A-Level Maths Mechanics - Question 3 - 2009 - Paper 1
Question 3
Two particles A and B are moving on a smooth horizontal plane. The mass of A is km, where 2 < k < 3, and the mass of B is m. The particles are moving along the same ... show full transcript
Worked Solution & Example Answer:Two particles A and B are moving on a smooth horizontal plane - Edexcel - A-Level Maths Mechanics - Question 3 - 2009 - Paper 1
Step 1
Find, in terms of k and u, the speed of B immediately after the collision.
Answer
To find the speed of B after the collision, we will apply the principles of conservation of momentum.
Before the collision:
- The momentum of A = mass ( km ) ( \times ) speed ( 2u ) = ( 2km u )
- The momentum of B = mass ( m ) ( \times ) speed ( -4u ) = ( -4mu )
Total momentum before collision = ( 2km u - 4mu )
After the collision, the speed of A is halved and its direction reversed, so:
- Speed of A = ( -u )
- Momentum of A after collision = ( km \times (-u) = -km u )
Let the speed of B after the collision be ( v ), then:
- The momentum of B = ( mv )
Total momentum after collision = ( -km u + mv )
Setting these equal (momentum before = momentum after): [ 2km u - 4mu = -km u + mv ] Rearranging gives: [ mv = 3km u - 4mu ] [ v = \frac{3km u - 4mu}{m} ] [ v = (3k - 4)u ]
Step 2
State whether the direction of motion of B changes as a result of the collision, explaining your answer.
Answer
To determine if the direction of motion of B changes, we look at the calculated speed behavior.
We know:
- Initially, B moves with speed ( 4u ) in the positive direction (let’s take this as right).
- The speed change is given by ( (3k - 4)u ).
From the constraint ( k > 2 ) (given in the question), we have:
[ 3k - 4 > 6 - 4 = 2 ]
This means ( v > 0 ), hence B continues in the same direction.
Therefore, the direction of motion of B does not change since its velocity remains positive.
Step 3
Find, in terms of m and u, the magnitude of the impulse that A exerts on B in the collision.
Answer
The impulse exerted on B can be calculated using the change in momentum of B.
The change in momentum of B is given by: [ \Delta p = p_{final} - p_{initial} ]
Where:
- Initial momentum of B: ( p_{initial} = mu \times -4u = -4mu )
- Final momentum of B: ( p_{final} = mv = m(3k - 4)u )
Thus, the impulse is: [ I = m(3k - 4)u - (-4mu) ] [ I = m(3k - 4 + 4)u = m(3k)u ] Substituting ( k = \frac{7}{3} ): [ I = m(3 \times \frac{7}{3})u = 7mu ]
Therefore, the magnitude of the impulse that A exerts on B is ( 7mu ).