In the \"before\" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffi

Question

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Explanation / Answer

Frictional force on Car A during sliding = mg;
Hence, Frictional retardation of car A by friction during sliding (a) = mg/m = g = 0.15*9.8 = 1.47 m/sec2
Distance travelled before A stops = 6.1m
Hence velocity of A at start of sliding va= (2ada) = 4.235 m/sec

Similarly Frictional retardation of car B by friction during sliding (b) = g = 0.15*9.8 = 1.47 m/sec2
Distance travelled before B stops = 4.4 m
Hence velocity of B at start of sliding vb= (2bdb) = 3.597 m/sec

If speed of B before collision = v
Net momentum before collision = mbv
Net momentum after collision = mava + mbvb
Using conservation of linear momentum we have : mbv = mava + mbvb
v = mava + mbvb/ mb = (1100*4.235 + 1400*3.597)/1400
Hence velocity of B before collision v= 6.9245 m/sec

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