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 Newtons Second Law: 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

96%

114 rated

Answer

To find the speed of particle A after the collision, we can use the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.

Let the speed of A after the collision be denoted as $v$ . The initial momenta of A and B can be calculated as follows:

Momentum of A before collision: ( p_A = m_A \cdot v_A = 0.4 \cdot 6 = 2.4 , \text{kg m s}^{-1} )
Momentum of B before collision: ( p_B = m_B \cdot v_B = 0.3 \cdot (-2) = -0.6 , \text{kg m s}^{-1} ) (since B is moving in the opposite direction)

Now, we can write the conservation of momentum equation:

$ext{Total Momentum before} = ext{Total Momentum after$ $2.4 - 0.6 = 0.4 \cdot v + 0.3 \cdot 3$

Simplifying gives us:

$2.4 - 0.6 = 0.4v + 0.9$ $1.8 = 0.4v + 0.9$ $0.9 = 0.4v$

Thus: $v = \frac{0.9}{0.4} = 2.25 \, \text{m s}^{-1}$

The direction of motion of A has not changed as the calculated speed remains 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

99%

104 rated

Answer

Impulse can be defined as the change in momentum. To calculate the impulse exerted on B, we use the following formula:

$I = m_B(v_{B_{final}} - v_{B_{initial}})$

Given:

Mass of B, ( m_B = 0.3 , \text{kg} )
Initial speed of B, ( v_{B_{initial}} = -2 , \text{m s}^{-1} ) (negative due to direction)
Final speed of B, ( v_{B_{final}} = 3 , \text{m s}^{-1} )

Now substituting the values in:

$I = 0.3 \cdot (3 - (-2))$ $I = 0.3 \cdot (3 + 2)$ $I = 0.3 \cdot 5 = 1.5 \, \text{Ns}$

Therefore, the magnitude of the impulse exerted on B in the collision is ( 1.5 , \text{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

Join the A-Level students using SimpleStudy...