Question

A truck with a mass of 1710 kg and moving with a speed of 14.0 m/s rear-ends a 613 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second. m/s truck m/s car

Answer 1

Mass of truck (mi) = 1710kg speed of truck (A) = 14 mis before eallifon speed of truck after v Calligan mass of ear = 613 ebed of ear before - M₂=0. speedo 1. Calligan speed of ear after eallison = V₂ : according to the law of confervation of momentum I. . mill, + m2 M 2 = m. v,+ m2 V2 1H0 x 14 +0 = 17100, +613 Vz. : 1710 V, +613 V, = 23940.-) The eallison is abbox elgtie so e=1. e=v2-v, an V2-v -n. 4,- 42 1 4 an, V-V, = 14 - Gi) forening (i) & (ii) we get, Vi= 6.611 mis 12 = 20.611 mis. Vear = Vq = 20.611 m/s V truck = V1 = 6.611 m/s.

b) m, has enery at the top = mgutt miya = 1.5X9.8 0,250 + 4x15x42 velverts just before Callison of hy is = u 1 mm,²= noch might f mine B i = [2ght my va 2x9.8x0.250 + 42 =4.57 m/s. just after callison Velocity of my is = 4.57m/s. Velocity of my ý=o. Conforming the linear marmentum mim, tm₂ M₂ = m, vit m₂ V2 [u=0]. 1.5X4.57 +0 = 1.5V, +4.7 V2 . 1.5V, +4.7V2 = 6-855 -(i) the Calliyan is eleglic so. e=1. u-M V2-V=4.57 -60) salning (i) & (ii) we get I. VE-2.36 my s. V2 = 2.21 ms. :. mi will more to the abboyite direction. 1 min?- mign, an, non un o E (2.36) = 0.284 m Fou my t har 12796 0.25 m m, will guing at a height of mz ? " " " " " 0.28am. 0.25 m.