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1000 Kg car approaches an intersection traveling north at 20 m/s . A 1150 Kg car approaches the same intersection traveling east at 33 m/s .
Question
1000 Kg car approaches an intersection traveling north at 20 m/s . A 1150 Kg car approaches the same intersection traveling east at 33 m/s . The two cars collide at the intersection and lock together. Ignoring any external forces that act on the cars during the collision, what is the velocity of the cars immediately after the collision
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Physics
5 years
2021-08-04T14:08:59+00:00
2021-08-04T14:08:59+00:00 1 Answers
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Answers ( )
Answer:
Explanation:
We shall represent direction of east by unit vector i and direction of north by vector j
momentum of 1000 kg car = mv j
= 1000 x 20 j
20000 j
momentum of 1150 kg car = mvi
= 1150 x 33 i
= 37950 i
Total momentum
= 37950 i +20000 j
Let their common velocity be v after collision
momentum after collision
= (1000 + 1150 )v
= 2150v
applying conservation of momentum law
2150 v = 37950 i +20000 j
v = 17.65 i + 9.30 j
magnitude of velocity
= √(17.65² + 9.3²)
= 19.95 m / s
direction θ with east
tanθ = 9.3 / 17.65
θ = 28 degree north of east .