A particle P of mass 0.3 kg is moving with speed u m s⁻¹ in a straight line on a smooth horizontal table. The particle P collides directly with a particle Q of mass 0.6 kg, which is at rest on the table. Immediately after the particles collide, P has speed 2 m s⁻¹ and Q has speed 5 m s⁻¹. The direction of motion of P is reversed by the collision. Find (a) the value of u, (b) the magnitude of the impulse exerted by P on Q. Immediately after the collision, a constant force of magnitude R newtons is applied to Q in the direction directly opposite to the direction of motion of Q. As a result Q is brought to rest in 1.5 s. (c) Find the value of R.

A-Level Edexcel Maths Mechanics Forces: A particle P of mass 0.3 kg is m

A particle P of mass 0.3 kg is moving with speed u m s⁻¹ in a straight line on a smooth horizontal table - Edexcel - A-Level Maths Mechanics - Question 4 - 2007 - Paper 1

Question 4

A particle P of mass 0.3 kg is moving with speed u m s⁻¹ in a straight line on a smooth horizontal table. The particle P collides directly with a particle Q of mass ... show full transcript

Worked Solution & Example Answer:A particle P of mass 0.3 kg is moving with speed u m s⁻¹ in a straight line on a smooth horizontal table - Edexcel - A-Level Maths Mechanics - Question 4 - 2007 - Paper 1

Step 1

Find the value of u.

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Answer

To find the value of u, we apply the principle of conservation of linear momentum (CLM).

The initial momentum of the system is equal to the final momentum:

$0.3u + 0.6 \times 0 = 0.3 \times (-2) + 0.6 \times 5$

Solving this equation:

Left Side: (0.3u = 0.3 \times (-2) + 0.6 \times 5)
Which simplifies to: (0.3u = -0.6 + 3)
Therefore: (0.3u = 2.4)
Dividing both sides by 0.3 gives (u = 8) m s⁻¹.

Step 2

Find the magnitude of the impulse exerted by P on Q.

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Answer

The impulse exerted by P on Q can be calculated using the formula:

$I = \Delta p = m \Delta v$

Where m is the mass of Q and (\Delta v) is the change in velocity.

Here:

Mass of Q, m = 0.6 kg
Change in velocity, (\Delta v = 5 - 0 = 5)

Thus:

$I = 0.6 \times 5 = 3\, \text{Ns}$

Step 3

Find the value of R.

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Answer

To find R, we first use the equation relating final velocity, initial velocity, acceleration, and time:

$v = u + at$

Here, Q comes to rest, so final velocity v = 0, u = 5, and time t = 1.5 s:

0 = 5 + a \times 1.5, a = \frac{-5}{1.5} = -\frac{10}{3} \text{ m s}^{-2}.

Next, using Newton's second law (N2L):

$F = ma$

The force applied to Q is R = 0.6 kg (\times -\frac{10}{3} m s^{-2} = -2, \text{N}$$

Thus, R = 2 N.

A particle P of mass 0.3 kg is moving with speed u m s⁻¹ in a straight line on a smooth horizontal table - Edexcel - A-Level Maths Mechanics - Question 4 - 2007 - Paper 1

Worked Solution & Example Answer:A particle P of mass 0.3 kg is moving with speed u m s⁻¹ in a straight line on a smooth horizontal table - Edexcel - A-Level Maths Mechanics - Question 4 - 2007 - Paper 1

Find the value of u.

Find the magnitude of the impulse exerted by P on Q.

Find the value of R.

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