A 750 kg car is hit head on by a 1500 kg truck. Before the collision the car was

Question

A 750 kg car is hit head on by a 1500 kg truck. Before the collision the car was moving with a velocity of 25 m/s North and the truck was moving South with a velocity of 35 m/s South. After the collision the truck moved South with a velocity of 21 m/s and moved West with a velocity of 8 m/s. A) What is the total momentum of the two vehicles before the collision? You will need 2 parts to the answer, one for the South direction and one for the West direction. WATCH DIRECTION! B) What is the total momentum of the two vehicles AFTER the collision? You will need 2 parts to the answer, one for the South direction and one for the West direction. WATCH DIRECTION! C) What is the vector velocity of the car immediately after the collision? D) What type of collision was this, elastic or inelastic? Explain how you determined this.

Explanation / Answer

here,

mass of truck , mt = 750 kg

mass of car , mc = 1500 kg

initial speed of car , uc = 25 m/s j

initial speed of truck , ut = - 35 m/s j

final speed of truck , vt = ( - 8 m/s i - 21 m/s j )

A)

the total momentum of two vehicle before the collison , Pi = mc*uc + mt*ut

Pi = 750 * 25 j - 1500 * 35 j

Pi = - 33750 kg.m/s j

the initial momentum is 33750 kg.m/s towards south

B)

as the momentum remains conserved

the final momentum = initial momentum

final momentum is 33750 kg.m/s towards south

C)

let the vector velocity of car immediately after the collison be vc

mt * vt + mc * vc = Pf

1500 * ( - 8 m/s i - 21 m/s j ) + 750 * vc = - 33750 j

vc = 16 i m/s - 3 m/s j

the final velocity of car after the collison is 16 m/s towards east and 3 m/s towards south

D)

initial kinetic energy , KEI = 0.5 * ( mc * uc^2 +mt * ut^2)

KEi = 1153125 J

the final kinetic energy , KEf = 0.5 * ( mc * vc^2 + mt * vt^2)

KEf = 478065 J

th initial kinetic energy is not equal to final

so, the collison is inelastic

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