At what speed (m/s) would a 1140-kg car have the same momentum as a 12.600-kg tr

Question

At what speed (m/s) would a 1140-kg car have the same momentum as a 12.600-kg truck travelling at 20.3 km/hr? The car and truck in (a) have a head-on collision and then stick together. What is their final common velocity (m/s)? (Assume the car is going in the positive direction.) What is the car's change in momentum in (b) (kg m/s)? At what speed would the same car have the same kinetic energy as the truck? The car and truck in (d) have a head-on collision and then stick together. What is their final common velocity (m/s)? (Assume the car is going in the positive direction.) What is the car's change in momentum in (e) (kg m/s)?

Explanation / Answer

a)

m1v1 = m2v2

1140*v = 12600*20.3 km/hr

v = 12600*20.3/1140 = 224.368 km/hr = 62.324 m/s

b)

we have total momentum is conserved

intial momentum = final momentum

m1*u1+m2*u2 = (m1+m2)*v

1140*224.368+12600*(-20.3) = (1140+12600)*v

v = 0

fina velocity is zero

c)

change in car's momentum = intial-final

= 1140*224.368-0 = 255779.52 KgKm/hr

or = 1140*62.324 -0 = 71049.36 Kgm/s

d)

1/2*m1v1^2 = 1/2*m2*v2^2

1140*v1^2 = 12600*20.3^2 km/hr

v^2 = 12600*20.3^2/1140

v = 67.488 km/hr = 18.746 m/s

e)

we have total momentum is conserved

intial momentum = final momentum

m1*u1+m2*u2 = (m1+m2)*v

1140*67.488+12600*(-20.3) = (1140+12600)*v

v = -13.016 km/hr = -3.615 m/s

f)

change in car's momentum = intial-final

= 1140*18.746-1140*(-3.615) = 25491.54 kgm/s

Understand the concept please dont mind the calculations they may be wrong so please do it again

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