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 Forces: 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

Find 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 A after the collision, we use the law of conservation of momentum:

$m_A v_A + m_B v_B = m_A v_A' + m_B v_B'$

Let:

Mass of A, $m_A = 0.4$ kg
Mass of B, $m_B = 0.3$ kg
Initial speed of A, $v_A = 6$ m/s (positive)
Initial speed of B, $v_B = -2$ m/s (negative, as it moves in the opposite direction)
Final speed of A, $v_A'$
Final speed of B, $v_B' = 3$ m/s (positive, as it is reversed)

Substituting the known values:

$0.4 imes 6 + 0.3 imes (-2) = 0.4 v_A' + 0.3 imes 3$

Calculating the left side:

$2.4 - 0.6 = 0.4 v_A' + 0.9$

Simplifying further:

$1.8 = 0.4 v_A' + 0.9$

Now, isolate $v_A'$ :

$1.8 - 0.9 = 0.4 v_A'$

$0.9 = 0.4 v_A'$

$v_A' = rac{0.9}{0.4} = 2.25 ext{ m/s}$

Since the direction of motion of A was not influenced (it continues in the same direction), we conclude that the speed of A immediately after the collision is 2.25 m/s, with its direction unchanged.

Step 2

Find 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 is calculated as the change in momentum. The formula for impulse ( $I$ ) is given by:

$I = m_B(v_B' - v_B)$

where:

$m_B = 0.3$ kg
$v_B' = 3$ m/s (final speed)
$v_B = -2$ m/s (initial speed)

Substituting the values:

$I = 0.3 imes (3 - (-2))$

$= 0.3 imes (3 + 2)$

$= 0.3 imes 5$

$= 1.5 ext{ kg m/s}$

Thus, the magnitude of the impulse exerted on B in the collision is 1.5 kg m/s.

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

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

Find 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...