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 straight line, but in opposite directions, and they collide directly. Immediately before they collide the speed of A is 2u and the speed of B is 4u. As a result of the collision the speed of A is halved and its direction of motion is reversed. (a) Find, in terms of k and u, the speed of B immediately after the collision. (b) State whether the direction of motion of B changes as a result of the collision, explaining your answer. Given that k = \frac{7}{3}, (c) find, in terms of m and u, the magnitude of the impulse that A exerts on B in the collision.

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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

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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.

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Answer

To find the speed of B immediately after the collision, we can use the principle of conservation of momentum.

Let the speed of B after the collision be v. Initially, the momentum before collision is:

$ext{Momentum before} = km(2u) + m(-4u) = 2km u - 4mu$

After the collision, the momentum is:

$ext{Momentum after} = km(-u) + mv$

Setting the total momentum before and after the collision equal gives:

$2km u - 4mu = -km u + mv$

Rearranging leads to:

$km(2u + u) = mv + 4mu$

This simplifies to:

$km(3u) = mv + 4mu$

Now, solving for v:

$v = rac{km(3u) - 4mu}{m} = 3ku - 4u$

Thus, substituting for k = \frac{7}{3}, we find:

$v = 3 \cdot \frac{7}{3}u - 4u = 7u - 4u = 3u.$

Step 2

State whether the direction of motion of B changes as a result of the collision, explaining your answer.

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Answer

The direction of the motion of B depends on the calculated speed after the collision. From the previous calculation, we found:

$v = 3u.$

Since B was initially traveling in the positive direction (4u), and after the collision its speed remains positive, we conclude that:

The direction of motion of B does not change as a result of the collision. Its speed remains in the same direction.

Step 3

Find, in terms of m and u, the magnitude of the impulse that A exerts on B in the collision.

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Answer

The impulse is defined as the change in momentum. The momentum of particle B before the collision is:

$p_{B, ext{before}} = mv_{B, ext{initial}} = m(-4u) = -4mu.$

After the collision, as calculated previously, the momentum of particle B is:

$p_{B, ext{after}} = mv = 3mu.$

Thus, the change in momentum for B is:

$\Delta p_{B} = p_{B, ext{after}} - p_{B, ext{before}} = 3mu - (-4mu) = 3mu + 4mu = 7mu.$

Consequently, the magnitude of the impulse that A exerts on B during the collision is:

$|J| = 7mu.$

Two particles A and B are moving on a smooth horizontal plane - Edexcel - A-Level Maths Mechanics - Question 3 - 2009 - Paper 1

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

Find, in terms of k and u, the speed of B immediately after the collision.

State whether the direction of motion of B changes as a result of the collision, explaining your answer.

Find, in terms of m and u, the magnitude of the impulse that A exerts on B in the collision.

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