Dynamics: Question 8

Syllabus 3.1, 3.2, 3.3

Structured AS 8 marks

Trolley CC, of mass 0.20 kg0.20\text{ kg}, travels at 6.0 m s16.0\text{ m s}^{-1} along a frictionless horizontal air track and collides with trolley DD, of mass 0.40 kg0.40\text{ kg}, which is initially at rest on the same track. The collision between the trolleys is perfectly elastic.

(a) State the two conditions that must both be satisfied for a collision to be described as perfectly elastic. [2]

(b) For a perfectly elastic collision, the relative speed of approach of the two bodies equals their relative speed of separation. Use this fact, together with conservation of momentum, to determine the velocity of each trolley immediately after the collision. Take the direction of trolley CC's initial motion as positive. [4]

(c) By calculating the total kinetic energy of the system immediately before and immediately after the collision, verify that the collision is indeed perfectly elastic. [2]

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Worked solution

Part (a): Conditions for a perfectly elastic collision

A collision is perfectly elastic only if both of the following are true:

  • the total linear momentum of the system is conserved, and
  • the total kinetic energy of the system is conserved (unchanged by the collision).

(Momentum alone is conserved in every collision, elastic or not. It is the additional conservation of kinetic energy that makes a collision perfectly elastic.)

Part (b): Velocities after the collision

Let vCv_C and vDv_D be the velocities of trolleys CC and DD immediately after the collision, with CC‘s initial direction taken as positive.

Conservation of momentum: mCuC+mDuD=mCvC+mDvDm_C u_C + m_D u_D = m_C v_C + m_D v_D (0.20)(6.0)+(0.40)(0)=0.20vC+0.40vD(0.20)(6.0) + (0.40)(0) = 0.20\,v_C + 0.40\,v_D 1.2=0.20vC+0.40vD...(1)1.2 = 0.20\,v_C + 0.40\,v_D \quad \text{...(1)}

Equal approach and separation speeds: the relative speed of approach is uCuD=6.00=6.0 m s1u_C-u_D=6.0-0=6.0\text{ m s}^{-1}, and for a perfectly elastic collision this equals the relative speed of separation, vDvCv_D-v_C: vDvC=6.0...(2)v_D - v_C = 6.0 \quad \text{...(2)}

Solve simultaneously. From (2): vD=vC+6.0v_D = v_C + 6.0. Substitute into (1): 1.2=0.20vC+0.40(vC+6.0)=0.60vC+2.41.2 = 0.20\,v_C + 0.40(v_C+6.0) = 0.60\,v_C + 2.4 0.60vC=1.22.4=1.2    vC=2.0 m s10.60\,v_C = 1.2-2.4 = -1.2 \implies v_C = -2.0\text{ m s}^{-1} vD=2.0+6.0=4.0 m s1v_D = -2.0+6.0 = 4.0\text{ m s}^{-1}

So trolley CC rebounds at 2.0 m s12.0\text{ m s}^{-1} in the reverse direction, while trolley DD moves off at 4.0 m s14.0\text{ m s}^{-1} in CC‘s original direction of motion.

Consistency check: momentum after =0.20×(2.0)+0.40×4.0=0.4+1.6=1.2 kg m s1=0.20\times(-2.0)+0.40\times4.0=-0.4+1.6=1.2\text{ kg m s}^{-1}, matching the momentum before the collision.

Part (c): Verifying the collision is elastic

Kinetic energy before the collision (only CC moving): Ek,before=12(0.20)(6.0)2=12(0.20)(36)=3.6 JE_{k,\text{before}} = \tfrac12(0.20)(6.0)^2 = \tfrac12(0.20)(36) = 3.6\text{ J}

Kinetic energy after the collision: Ek,after=12(0.20)(2.0)2+12(0.40)(4.0)2=12(0.20)(4.0)+12(0.40)(16)E_{k,\text{after}} = \tfrac12(0.20)(-2.0)^2 + \tfrac12(0.40)(4.0)^2 = \tfrac12(0.20)(4.0)+\tfrac12(0.40)(16) =0.4+3.2=3.6 J= 0.4+3.2 = 3.6\text{ J}

Since Ek,after=Ek,before=3.6 JE_{k,\text{after}}=E_{k,\text{before}}=3.6\text{ J}, kinetic energy is conserved, confirming the collision is indeed perfectly elastic.

Final answers

  • (a) Momentum conserved and kinetic energy conserved
  • (b) vC=2.0 m s1v_C = \boxed{-2.0}\text{ m s}^{-1} (rebounds), vD=4.0 m s1v_D = \boxed{4.0}\text{ m s}^{-1} (forward)
  • (c) Ek,before=Ek,after=3.6 JE_{k,\text{before}}=E_{k,\text{after}}=\boxed{3.6}\text{ J}, perfectly elastic confirmed