Three gliders, A, B and C, move on a straight, smooth, horizontal air track. Glider A has mass 0.6 kg and moves at 4 m s^-1 towards glider B, which has mass 0.9 kg and is initially at rest. A and B collide directly; they do not coalesce. (a) Given that immediately after this fir…

Answer: ((a)) vB = 2.8 m s^-1 (in A's original direction of motion) ((b)) common velocity of B and C = 1.05 m s^-1 (in A's original direction of motion) ((c)) A moves in the negative direction (opposite to its original motion); since A and the combined B-and-C glider then move apart in opposite directions from the collision point, A can never catch up with them. Setting up the sign convention Take A's original direction of motion as positive throughout. Part (a): First collision, A with B Before the first collision, B is at rest, so uB = 0. Total momentum before: p1 = mA uA + mB uB = (0.6)(4) + (0…

A Level Maths 9709: Two-Stage Collision on an Air Track

Three gliders, $A$ , $B$ and $C$ , move on a straight, smooth, horizontal air track. Glider $A$ has mass $0.6\text{ kg}$ and moves at $4\text{ m s}^{-1}$ towards glider $B$ , which has mass $0.9\text{ kg}$ and is initially at rest. $A$ and $B$ collide directly; they do not coalesce.

(a) Given that immediately after this first collision $A$ has velocity $-0.2\text{ m s}^{-1}$ (that is, it rebounds), find the velocity of $B$ immediately after the collision. [3]

(b) Glider $B$ , now moving with the velocity found in part (a), goes on to collide with glider $C$ , mass $1.5\text{ kg}$ , which is at rest further along the track. $B$ and $C$ coalesce on impact. Find the common velocity of $B$ and $C$ immediately after this second collision. [3]

(c) State the direction in which $A$ moves after its collision with $B$ , and explain whether $A$ can ever catch up with the combined glider formed by $B$ and $C$ . [2]

Setting up the sign convention

Take $A$ ‘s original direction of motion as positive throughout.

Part (a): First collision, $A$ with $B$

Before the first collision, $B$ is at rest, so $u_B = 0$ . Total momentum before: $p_1 = m_A u_A + m_B u_B = (0.6)(4) + (0.9)(0) = 2.4\text{ kg m s}^{-1}$

By conservation of momentum, this total is unchanged by the collision. Using $A$ ‘s given velocity after the collision, $v_A = -0.2\text{ m s}^{-1}$ (the minus sign shows $A$ rebounds): $m_A v_A + m_B v_B = p_1$ $(0.6)(-0.2) + (0.9)v_B = 2.4$ $-0.12 + 0.9v_B = 2.4$ $0.9v_B = 2.52$ $v_B = \frac{2.52}{0.9} = 2.8\text{ m s}^{-1}$

So $B$ moves off at $2.8\text{ m s}^{-1}$ in $A$ ‘s original (positive) direction.

Part (b): Second collision, $B$ with $C$

This second collision involves only $B$ (now moving at $2.8\text{ m s}^{-1}$ ) and $C$ (at rest, $u_C = 0$ ). Total momentum of this pair before the second collision: $p_2 = m_B v_B + m_C u_C = (0.9)(2.8) + (1.5)(0) = 2.52\text{ kg m s}^{-1}$

Since $B$ and $C$ coalesce, they share one common velocity $v$ afterwards, with combined mass: $m_B + m_C = 0.9 + 1.5 = 2.4\text{ kg}$

By conservation of momentum: $p_2 = (m_B+m_C)v$ $2.52 = 2.4v$ $v = \frac{2.52}{2.4} = 1.05\text{ m s}^{-1}$

So $B$ and $C$ move off together at $1.05\text{ m s}^{-1}$ , still in $A$ ‘s original (positive) direction.

Part (c): Direction of $A$ , and whether it can catch up

From part (a), $A$ ‘s velocity after the first collision is $-0.2\text{ m s}^{-1}$ : since this is negative, $A$ moves in the direction opposite to its original motion (it rebounds backwards).

Meanwhile, the combined glider formed by $B$ and $C$ moves at $+1.05\text{ m s}^{-1}$ , in $A$ ‘s original direction. So from the moment of the first collision onward, $A$ travels one way along the track while $B$ -and- $C$ travel the other way, the two move apart from that point, with the gap between them increasing at a constant rate of $0.2+1.05=1.25\text{ m s}^{-1}$ . Since they never move back towards each other again, $A$ can never catch up with the combined glider $B$ -and- $C$ .

Final answers

(a) $v_B = \boxed{2.8}\text{ m s}^{-1}$ , in $A$ ‘s original direction of motion
(b) Common velocity of $B$ and $C$ $= \boxed{1.05}\text{ m s}^{-1}$ , in $A$ ‘s original direction of motion
(c) $A$ moves in the negative direction (opposite to its original motion); as it and $B$ -and- $C$ then move apart in opposite directions, $A$ can never catch up with them

Worked solution

Setting up the sign convention

Part (a): First collision, AAA with BBB

Part (b): Second collision, BBB with CCC

Part (c): Direction of AAA, and whether it can catch up

Final answers

Part (a): First collision, $A$ with $B$

Part (b): Second collision, $B$ with $C$

Part (c): Direction of $A$ , and whether it can catch up