Trolley A, of mass 0.40 kg, travels at 1.8 m s^-1 along a horizontal, frictionless air track and collides with trolley B, of mass 0.60 kg, which is initially at rest on the same track. The two trolleys have magnetic pads that lock together on impact, so immediately after the col…

Answer: ((a)) common velocity v = 0.72 m s^-1, in the direction of A's original motion ((b)) kinetic energy before = 0.648 J, kinetic energy after = 0.259 J (3 s.f.); since kinetic energy decreases, the collision is inelastic ((c)) no external resultant force acts on the isolated two-trolley system, so momentum is conserved regardless of collision type; kinetic energy is only conserved in elastic collisions, and here some is transformed into other forms of energy (e.g. sound, heat, deformation of the magne…. Part (a): Common velocity after the collision Take the direction of A's initial motion as positive. Since B starts at rest, the total momentum before the collision is: pbefore = mA uA + mB uB = (0.40)(1.8) + (0.60)(0) = 0.72 kg m s^-1 Sinc…

A Level Physics 9702: Colliding Trolleys and Conservation of Momentum

Trolley $A$ , of mass $0.40\text{ kg}$ , travels at $1.8\text{ m s}^{-1}$ along a horizontal, frictionless air track and collides with trolley $B$ , of mass $0.60\text{ kg}$ , which is initially at rest on the same track. The two trolleys have magnetic pads that lock together on impact, so immediately after the collision they move off with a single common velocity.

(a) Calculate the common velocity of the two trolleys immediately after the collision. [2]

(b) By calculating the total kinetic energy of the system immediately before and immediately after the collision, determine whether the collision is elastic or inelastic. [3]

(c) Explain why the linear momentum of the two-trolley system is conserved in this collision, even though kinetic energy is not conserved. [2]

Part (a): Common velocity after the collision

Take the direction of $A$ ‘s initial motion as positive. Since $B$ starts at rest, the total momentum before the collision is: $p_{\text{before}} = m_A u_A + m_B u_B = (0.40)(1.8) + (0.60)(0) = 0.72\text{ kg m s}^{-1}$

Since the trolleys lock together, they share one common velocity $v$ afterward, moving with combined mass $m_A + m_B = 0.40+0.60=1.0\text{ kg}$ . By conservation of momentum: $p_{\text{before}} = (m_A+m_B)v$ $0.72 = 1.0 \times v$ $v = 0.72\text{ m s}^{-1}$

The common velocity is $\boxed{0.72}\text{ m s}^{-1}$ , in the direction of $A$ ‘s original motion (since $v$ is positive).

Part (b): Kinetic energy before and after

Kinetic energy before the collision (only $A$ is moving, $B$ is at rest): $E_{k,\text{before}} = \tfrac{1}{2}m_A u_A^2 = \tfrac{1}{2}(0.40)(1.8)^2 = \tfrac{1}{2}(0.40)(3.24) = 0.648\text{ J}$

Kinetic energy after the collision (both trolleys move together at $v=0.72\text{ m s}^{-1}$ ): $E_{k,\text{after}} = \tfrac{1}{2}(m_A+m_B)v^2 = \tfrac{1}{2}(1.0)(0.72)^2 = \tfrac{1}{2}(1.0)(0.5184) = 0.2592\text{ J}$

Comparing the two values: $E_{k,\text{before}} = 0.648\text{ J} \qquad E_{k,\text{after}} = 0.259\text{ J (3 s.f.)}$

Since $E_{k,\text{after}} < E_{k,\text{before}}$ , kinetic energy has decreased by $0.648-0.259=0.389\text{ J}$ . Kinetic energy is not conserved, so the collision is inelastic. In fact, since the trolleys stick together and move as one body, this is the maximally inelastic case (a perfectly inelastic collision).

Part (c): Why momentum is conserved but kinetic energy is not

Momentum is conserved in any collision, elastic or inelastic, provided no external resultant force acts on the system. Here the track is frictionless and the collision forces between $A$ and $B$ are internal (equal and opposite, by Newton’s third law), they cancel out when considering the trolleys as one system, so the total momentum of the system is unchanged by the collision.

Kinetic energy, by contrast, is only guaranteed to be conserved in a perfectly elastic collision. In this collision the trolleys deform the magnetic pads and lock together, and some of the original kinetic energy is transformed into other forms of energy (such as heat, sound, and energy stored in the deformation of the pads) rather than remaining as kinetic energy of motion. Energy overall is still conserved; it is simply no longer all in the form of kinetic energy.

Final answers

(a) Common velocity $= \boxed{0.72}\text{ m s}^{-1}$ , in the direction of $A$ ‘s original motion
(b) $E_{k,\text{before}} = \boxed{0.648}\text{ J}$ , $E_{k,\text{after}} = \boxed{0.259}\text{ J}$ . The collision is inelastic
(c) Momentum is conserved because no external resultant force acts on the system; kinetic energy is not conserved because energy is transformed into other forms (heat, sound, deformation) during the collision