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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 apply the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision:
Initial momentum: [ km(2u) + m(-4u) = 2km u - 4mu ]
After the collision: [ km(v) + m(u) ]
Setting these equal gives: [ km(2u) - 4mu = km(v) + mu ]
Rearranging this results in: [ km(2u) - mu = km(v) + 4mu ]
We can further simplify to find the value of v: [ v(3k - 4) = u(3k - 4) ]
Thus, we find: [ v = \frac{u(3k - 4)}{k} ]
Step 2
State whether the direction of motion of B changes as a result of the collision, explaining your answer.
Answer
The direction of motion of B will change if its final velocity is positive. Since we have established that the speed of B after the collision is proportional to ( 3k - 4 ), we can analyze the value of k. Given that ( k > 2 ) and since ( k < 3 ), we can evaluate: [ 3k - 4 > 0 ] implies [ k > \frac{4}{3} ]. Therefore, since ( k ) is in the range stated, B's direction does indeed reverse as a result of the collision.
Step 3
Find, in terms of m and u, the magnitude of the impulse that A exerts on B in the collision.
Answer
To calculate the impulse exerted by A on B, we utilize the impulse-momentum theorem. Impulse is defined as the change in momentum:
The initial momentum of B before the collision is ( m(-4u) ) and after it is ( mv ), so: [ , \text{Impulse} = mv - m(-4u) ] [ = m(v + 4u) ] Substituting our earlier finding for v we get: [ = m\left(\frac{u(3k - 4)}{k} + 4u\right) ] [ = m\left(\frac{u(3k - 4 + 4k)}{k}\right) ] [ = m\left(\frac{u(7k - 4)}{k}\right) ] When substituting ( k = \frac{7}{3} ), we find the impulse exerted by A on B.