Two particles A and B have mass 0.4 kg and 0.3 kg respectively. They are moving in opposite directions on a smooth horizontal table and collide directly. Immediately before the collision, the speed of A is 6 m s⁻¹ and the speed of B is 2 m s⁻¹. As a result of the collision, the direction of motion of B is reversed and its speed immediately after the collision is 3 m s⁻¹. Find a) the speed of A immediately after the collision, stating clearly whether the direction of motion of A is changed by the collision, b) the magnitude of the impulse exerted on B in the collision, stating clearly the units in which your answer is given.

A-Level Edexcel Maths Mechanics Further Forces & Newtons Laws: Two particles A and B have mass

Two particles A and B have mass 0.4 kg and 0.3 kg respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2006 - Paper 1

Question 2

Two particles A and B have mass 0.4 kg and 0.3 kg respectively. They are moving in opposite directions on a smooth horizontal table and collide directly. Immediately... show full transcript

Worked Solution & Example Answer:Two particles A and B have mass 0.4 kg and 0.3 kg respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2006 - Paper 1

Step 1

a) the speed of A immediately after the collision, stating clearly whether the direction of motion of A is changed by the collision

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Answer

To determine the speed of particle A after the collision, we use the principle of conservation of momentum. The initial momentum of the system before the collision can be calculated as:

$P_{initial} = m_A v_A + m_B v_B$

where:

$m_A = 0.4 \, kg$ , $v_A = 6 \, m/s$
$m_B = 0.3 \, kg$ , $v_B = -2 \, m/s$ (negative because it moves in the opposite direction)

Substituting in the values:

$P_{initial} = (0.4 \, kg)(6 \, m/s) + (0.3 \, kg)(-2 \, m/s)$

Calculating:

$P_{initial} = 2.4 - 0.6 = 1.8 \, kg\cdot m/s$

After the collision, momentum is conserved. Let $v_A$ be the speed of A after the collision, then:

$P_{final} = m_A v_A + m_B v_B'$

where $v_B' = 3 \, m/s$ . Using conservation of momentum:

$1.8 = (0.4)v_A + (0.3)(3)$

Simplifying this:

$1.8 = 0.4 v_A + 0.9$

Subtracting 0.9 from both sides:

$0.9 = 0.4 v_A$

Finally, solving for $v_A$ :

$v_A = rac{0.9}{0.4} = 2.25 \, m/s$

The direction of motion of A does not change since its speed is positive.

Step 2

b) the magnitude of the impulse exerted on B in the collision, stating clearly the units in which your answer is given

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Answer

Impulse is defined as the change in momentum. The impulse exerted on B can be calculated as:

$I = m_B (v_B' - v_B)$

where $v_B' = 3 \, m/s$ (after the collision) and $v_B = -2 \, m/s$ (before the collision). Substituting the values:

$I = (0.3)(3 - (-2)) = (0.3)(3 + 2) = (0.3)(5) = 1.5 \, Ns$

Thus, the magnitude of the impulse exerted on B is 1.5 Ns.

Two particles A and B have mass 0.4 kg and 0.3 kg respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2006 - Paper 1

Worked Solution & Example Answer:Two particles A and B have mass 0.4 kg and 0.3 kg respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2006 - Paper 1

a) the speed of A immediately after the collision, stating clearly whether the direction of motion of A is changed by the collision

b) the magnitude of the impulse exerted on B in the collision, stating clearly the units in which your answer is given

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