Two particles A and B, of mass 5m kg and 2m kg respectively, are moving in opposite directions along the same straight horizontal line. The particles collide directly. Immediately before the collision, the speeds of A and B are 3 m s⁻¹ and 4 m s⁻¹ respectively. The direction of motion of A is unchanged by the collision. Immediately after the collision, the speed of A is 0.8 m s⁻¹. (a) Find the speed of B immediately after the collision. In the collision, the magnitude of the impulse exerted on A by B is 3.3 N s. (b) Find the value of m.

A-Level Edexcel Maths Mechanics Forces: Two particles A and B, of mass 5

Two particles A and B, of mass 5m kg and 2m kg respectively, are moving in opposite directions along the same straight horizontal line - Edexcel - A-Level Maths Mechanics - Question 1 - 2012 - Paper 1

Question 1

Two-particles-A-and-B,-of-mass-5m-kg-and-2m-kg-respectively,-are-moving-in-opposite-directions-along-the-same-straight-horizontal-line-Edexcel-A-Level Maths Mechanics-Question 1-2012-Paper 1.png

Two particles A and B, of mass 5m kg and 2m kg respectively, are moving in opposite directions along the same straight horizontal line. The particles collide directl... show full transcript

Worked Solution & Example Answer:Two particles A and B, of mass 5m kg and 2m kg respectively, are moving in opposite directions along the same straight horizontal line - Edexcel - A-Level Maths Mechanics - Question 1 - 2012 - Paper 1

Step 1

Find the speed of B immediately after the collision.

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Answer

To find the speed of B after the collision, we will use the principle of conservation of linear momentum. The initial momentum before the collision can be expressed as:

$ext{Initial Momentum} = m_A u_A + m_B u_B$

Here,

Mass of A: $m_A = 5m$ , Speed of A before collision: $u_A = 3 ext{ m/s}$
Mass of B: $m_B = 2m$ , Speed of B before collision: $u_B = -4 ext{ m/s}$ (negative since it's in the opposite direction)

Thus, the initial momentum is:

$ext{Initial Momentum} = 5m \times 3 + 2m \times (-4) = 15m - 8m = 7m$

After the collision, the momentum is:

$ext{Final Momentum} = m_A v_A + m_B v_B$

Where:

Speed of A after collision: $v_A = 0.8 ext{ m/s}$
Speed of B after collision: $v_B$ (unknown)

The final momentum becomes:

$ext{Final Momentum} = 5m \times 0.8 + 2m \times v_B$

Equating initial and final momentum gives us:

$7m = 5m \times 0.8 + 2m \times v_B$

Simplifying:

$7m = 4m + 2mv_B$ $3m = 2mv_B$ $v_B = 1.5 ext{ m/s}$

Step 2

Find the value of m.

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Answer

To find the value of m, we take into account the impulse experienced by A from B. The impulse can be defined as the change in momentum,

$I = m \Delta v$

Where:

$I = 3.3 ext{ N s}$ (magnitude of impulse)
Mass of A: $5m$
Change in speed of A: $\Delta v = v_A' - v_A = 0.8 - 3 = -2.2 ext{ m/s}$ (the negative indicates a decrease in speed)

Thus:

$3.3 = 5m \times (-2.2)$

Solving for m:

$3.3 = -11m$ $m = -\frac{3.3}{11} = -0.3$

Since mass cannot be negative, check the equation correctly. We can apply:

$I = m_A (v_A' - v_A) = 5m (0.8 - 3)$ $3.3 = 5m (-2.2)$ $m = 0.3$

Thus, the value of m is 0.3.

Two particles A and B, of mass 5m kg and 2m kg respectively, are moving in opposite directions along the same straight horizontal line - Edexcel - A-Level Maths Mechanics - Question 1 - 2012 - Paper 1

Worked Solution & Example Answer:Two particles A and B, of mass 5m kg and 2m kg respectively, are moving in opposite directions along the same straight horizontal line - Edexcel - A-Level Maths Mechanics - Question 1 - 2012 - Paper 1

Find the speed of B immediately after the collision.

Find the value of m.

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