A railway truck P, of mass m kg, is moving along a straight horizontal track with speed 15 ms⁻¹. A truck P collides with a truck Q of mass 3000 kg, which is at rest on the same track. Immediately after the collision the speed of P is 3 ms⁻¹ and the speed of Q is 9 ms⁻¹. The direction of motion of P is reversed by the collision. Modelling the trucks as particles, find a) the magnitude of the impulse exerted by P on Q. b) the value of m.

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A railway truck P, of mass m kg, is moving along a straight horizontal track with speed 15 ms⁻¹ - Edexcel - A-Level Maths Mechanics - Question 1 - 2012 - Paper 1

Question 1

A railway truck P, of mass m kg, is moving along a straight horizontal track with speed 15 ms⁻¹. A truck P collides with a truck Q of mass 3000 kg, which is at rest... show full transcript

Worked Solution & Example Answer:A railway truck P, of mass m kg, is moving along a straight horizontal track with speed 15 ms⁻¹ - Edexcel - A-Level Maths Mechanics - Question 1 - 2012 - Paper 1

Step 1

a) the magnitude of the impulse exerted by P on Q

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Answer

To find the magnitude of the impulse exerted by truck P on truck Q, we use the impulse-momentum theorem. The impulse (I) is given by:

$I = ext{Change in momentum of } Q$

Since truck Q was initially at rest, its initial momentum is 0. After the collision, the momentum of Q is:

$ext{Momentum of } Q = m_Q imes v_Q = 3000 ext{ kg} imes 9 ext{ ms}^{-1} = 27000 ext{ kg m/s}$

Thus, the magnitude of the impulse is:

$I = 27000 ext{ Ns}$

Step 2

b) the value of m

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Answer

For part (b), we apply the principle of conservation of linear momentum.

The total momentum before the collision equals the total momentum after the collision:

$m_P imes v_P + m_Q imes v_Q = m_P' imes v_P' + m_Q' imes v_Q'$

Substituting the known values, we have:

$m imes 15 - 3000 imes 0 = m imes (-3) + 3000 imes 9$

Rearranging the equation:

$15m = -3m + 27000$

Combining like terms gives:

$18m = 27000$

Thus, solving for m:

$m = \frac{27000}{18} = 1500 ext{ kg}$

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