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Momentum: Question 8

Syllabus 4.3

Structured AS 7 marks

At a fairground, bumper car CC (mass 250 kg250\text{ kg} with its rider) moves at 2 m s−12\text{ m s}^{-1} along a straight track. Bumper car DD (mass 300 kg300\text{ kg} with its rider) moves at 1.5 m s−11.5\text{ m s}^{-1} directly towards CC along the same track. The cars collide directly and do not coalesce.

(a) Taking the direction of CC's initial motion as positive, write down the signed initial velocities of CC and DD, and hence find the total momentum of the system before the collision. [1]

(b) Given that immediately after the collision CC has velocity −1 m s−1-1\text{ m s}^{-1} (that is, it rebounds), find the velocity of DD immediately after the collision. [4]

(c) State the direction each car moves in immediately after the collision, and explain whether the two cars can collide with each other again. [2]

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Worked solution

Part (a): Total momentum before the collision

Take the direction of CC‘s initial motion as positive. Car DD moves directly towards CC, i.e. in the opposite direction, so its velocity is negative: uC=2 m s−1,uD=−1.5 m s−1u_C = 2\text{ m s}^{-1}, \qquad u_D = -1.5\text{ m s}^{-1}

Total momentum before the collision: pbefore=mCuC+mDuD=(250)(2)+(300)(−1.5)=500−450=50 kg m s−1p_{\text{before}} = m_C u_C + m_D u_D = (250)(2) + (300)(-1.5) = 500 - 450 = 50\text{ kg m s}^{-1}

Part (b): Velocity of DD after the collision

Since no external horizontal forces act during the impact, momentum is conserved: the total momentum after the collision must still equal 50 kg m s−150\text{ kg m s}^{-1}.

Using CC‘s given velocity after the collision, vC=−1 m s−1v_C = -1\text{ m s}^{-1} (the minus sign shows CC rebounds): mCvC+mDvD=pbeforem_C v_C + m_D v_D = p_{\text{before}} (250)(−1)+(300)vD=50(250)(-1) + (300)v_D = 50 −250+300vD=50-250 + 300v_D = 50 300vD=300300v_D = 300 vD=1 m s−1v_D = 1\text{ m s}^{-1}

Part (c): Directions after the collision, and whether they collide again

CC‘s velocity after the collision is −1 m s−1-1\text{ m s}^{-1}: since this is negative, CC rebounds and moves in the direction opposite to its original motion, back the way DD originally came from.

DD’s velocity after the collision is +1 m s−1+1\text{ m s}^{-1}: since this is positive, DD now moves in the direction of CC‘s original motion, back the way CC originally came from.

So from the moment of collision, CC moves one way along the track and DD moves the other way, both heading away from the point where they met. Since they continue to separate and no further forces bring them back together, CC and DD cannot collide with each other again.

Final answers

  • (a) uC=2 m s−1u_C = \boxed{2}\text{ m s}^{-1}, uD=−1.5 m s−1u_D = \boxed{-1.5}\text{ m s}^{-1}; total momentum before =50 kg m s−1= \boxed{50}\text{ kg m s}^{-1}
  • (b) vD=1 m s−1v_D = \boxed{1}\text{ m s}^{-1}
  • (c) CC moves in the negative direction, DD moves in the positive direction; they move apart, so they cannot collide again