A railway truck P of mass 2000 kg is moving along a straight horizontal track with speed 10 m s⁻¹ - Edexcel - A-Level Maths Mechanics - Question 1 - 2003 - Paper 1

To find the speed of truck P after the collision, we can use the principle of conservation of momentum. The total momentum before the collision must equal the total momentum after the collision.

Let:

• mass of truck P, $m_P = 2000 \, \text{kg}$
• speed of truck P before the collision, $u_P = 10 \, \text{m s}^{-1}$
• mass of truck Q, $m_Q = 3000 \, \text{kg}$
• speed of truck Q before the collision, $u_Q = 0 \, \text{m s}^{-1}$ (since it is at rest)
• speed of truck P after the collision, $v_P$ (to be determined)
• speed of truck Q after the collision, $v_Q = 5 \, \text{m s}^{-1}$

Using the conservation of momentum:

$m_P u_P + m_Q u_Q = m_P v_P + m_Q v_Q$

Substituting in the values:

$2000 \times 10 + 3000 \times 0 = 2000 v_P + 3000 \times 5$

Simplifying gives:

$20000 = 2000 v_P + 15000$

Rearranging the equation to solve for $v_P$:

$2000 v_P = 20000 - 15000$ $2000 v_P = 5000$ $v_P = \frac{5000}{2000} = 2.5 \, \text{m s}^{-1}$

Thus, the speed of truck P immediately after the collision is 2.5 m s⁻¹.