Cars colliding on a barge - part II

Classical Mechanics Level 2

Prerequisites:

Accelerated motion: If a body moves with a constant acceleration a a , the distance covered by the body after a certain time t t is expressed as: s = a t 2 2 + v t s = \frac{at^2}{2}+vt and the body reaches the velocity: v = a t + v 0 , v = at + v_0, where v 0 v_0 is the initial velocity of the body.

The task

A barge of a mass m L m_L transports two cars A and B of masses m A m_A and m B m_B respectively. They are placed at the distance d d away from each other. They drive off with accelerations a A a_A and a B a_B respectively. Each one accelerates until it reaches a constant velocity v v with respect to the barge. After what time will the two cars collide?

Assumptions: The barge initially at rest. Any resistances are neglected. The cars are treated as mass points. Given: m L = 22500 m_L = 22500 kg, m A = 2000 m_A = 2000 kg, m B = 1500 m_B = 1500 kg, a A = 2 m s 2 a_A = 2\; \frac{\mathrm m}{\mathrm s^2} , a B = 4 m s 2 a_B = 4 \frac{\mathrm m}{\mathrm s^2} , v = 3 m s v = 3 \frac{\mathrm m}{\mathrm s} , d = 15 d = 15 m.


The answer is 3.0625.

Maria Paszkiewicz by Maria Paszkiewicz , 29, Germany

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1 solution

Maria Paszkiewicz
Jan 24, 2020

Considering a movement relative to the barge: a A t A = v a_At_A=v and a B t B = v a_Bt_B=v give the times t A = 1.5 s t_A=1.5\;\mathrm s for car A and t B = 0.75 s t_B=0.75\;\mathrm s for car B. It means that at the moment when car B already reached the velocity v v , car A still accelerates. The distance covered by car A up to the moment t A t_A can be calculated as:

s A ( t A ) = a A t A 2 2 = v 2 2 a A = 9 4 m . s_A(t_A)= \frac{a_At_A^2}{2}=\frac{v^2}{2a_A}= \frac{9}{4}\; \mathrm m.

The distance covered by car B by the same time t A t_A is:

s B ( t A ) = s B ( t B ) + v ( t A − t B ) = v 2 2 a B + v ( t A − t B ) = 27 8 m . s_B(t_A)=s_B(t_B)+v(t_A-t_B)=\frac{v^2}{2a_B}+v(t_A-t_B)= \frac{27}{8} \mathrm m.

The distance left for cars to collide is then:

x = d − s A ( t A ) − s B ( t A ) = 75 8 m . x = d-s_A(t_A)-s_B(t_A) = \frac{75}{8}\; \mathrm m.

Both cars have the same velocity $v$ after time t A t_A . Therefore they will collide at the point where each of them covered the distance x 2 \frac{x}{2} . Covering the distance x 2 \frac{x}{2} will take the time t x = x 2 v = 1.5625 s t_x=\frac{x}{2v}=1.5625 \mathrm s . So the total time after which the cars will collide equals:

t = t A + t x = 1.5 + 1.5625 = 3.0625 s . t = t_A + t_x=1.5 + 1.5625 = 3.0625\; \mathrm s.

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