(a) At what speed (m/s) would a 1140-kg car have the same momentum as a 12,600-k

Question

(a) At what speed (m/s) would a 1140-kg car have the same momentum as a 12,600-kg truck traveling at17.8km/hr?

(b) The car and truck in (a) have a head-on collision and then stick together. What is their nal common velocity (m/s)? (Assume the car is going in the positive direction.)

(c) What is the car's change in momentum in (b) (kg m/s)?

(d) At what speed would the same car have the same kinetic energy as the truck?

(e) The car and truck in (d) have a head-on collision and then stick together. What is their nal common velocity (m/s)? (Assume the car is going in the positive direction.)

(f) What is the car's change in momentum in (e) (kg m/s)?

The correct answers are (a)54.6, (b) 0, (c) -62300, (d) 16.4, (e) -3.17, (f) -22400

What are all the proper equations for each part?

Explanation / Answer

a)

m1v1 = m2v2

1140*v = 12600*17.8 => v = 196.74 km/hr = 54.56m/s

b)

As momentum is conserved in all collision, The common velocity will be 0 as both car and truck have equal and opposite momentum

c)

-m1v1

Car's momentum change = -1140*54.56 = -62300 kg m/s

d) 0.5*m1(v1)^2 = 0.5*m2(v2)^2

1140*v*v = 12600*17.8*17.8 => v = 59.18 km/hr = 16.44 m/s

e) Again by conservation of momentum:

velocity of truck in m/s = 17.8*1000/3600 = 4.94 m/s

m1v1+m2v2 = (m1+m2)v

1140*16.44-12600*4.94 = 13740*v = -3.17m/s

f) m1(v-v1) = 1140*(-3.17 - 16.44) = -22355.4 kg m/s

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