Question

Apps Canvas 3 Degree Req Class Plan + depop sales h Orders | HelpHome... Problem 1 A truck of mass 5000 kg traveling in the positive x-direction collides elastically with a car of mass 3000 kg initially at rest. After the collision, the truck moves at a known speed but in an unknown direction, while the car moves in a known direction but at an unknown speed. The details of the gentle collision are summarized in the diagam below. Over the course of the whole process, you may assume no net external force is acting on the system. Peruck = 4 m/s Before collision: Truck Car 2 m/s After collision: Truck Car 26.6° Ver A) (8 points) Determine the final speed Ucar of the car after the collision B) (12 points) Determine the direction of motion of the truck after the collision. It is labeled e in the diagram 9:29 AM 7/29/2020 Pa Home F9 End Pg UP Fn Prt Scn F8 1) F10 F6 F7 + & ( ) 7 8 9 0 o Р K L

Answer 1

Answers :

Part (a) Answer : v_car = 4.472 m/s

Part (b) Answer : θ = 36.9^o

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Solution :

Given :

Mass of truck (m_truck) = 5000 kg

mass of car (m_car) = 3000 kg

Initially :

v_truck = (4 m/s) i

v_car = 0 m/s

After collision :

v'_truck = {(2 m/s) cosθ} i + {(2 m/s) sinθ} j

v_car = {(v) cos(26.6)} i - {(v) sin(26.6)} j

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For elastic collision : Kinetic energy before collision will be equal to the kinetic energy after collision.

∴ (1/2)(m_truck)(v_truck)² + (1/2)(m_car)(v_car)² = (1/2)(m_truck)(v'_truck)² + (1/2)(m_car)(v_car)²

∴ (5000 kg)(4 m/s)² + (3000 kg)(0 m/s)² = (5000 kg)(2 m/s)² + (3000 kg)(v_car)²

∴ 80000 J = 20000 J + (3000 kg)(v_car)²

∴ 60000 J = (3000 kg)(v_car)²

∴ 20 m²/s² = (v_car)²

∴ v_car = 4.472 m/s

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Here initial momentum will be : P_initial = m_truck v_truck + m_car v_car

∴ P_initial = (5000 kg){(4 m/s) i} + (3000 kg)(0 m/s) = (20000 kg m/s) i

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And, Final momentum will be : P_final = m_truck v'_truck + m_car v_car

∴ P_final = (5000 kg) [ {(2 m/s) cosθ} i + {(2 m/s) sinθ} j ] + (3000 kg) [ {(v) cos(26.6)} i - {(v) sin(26.6)} j ]

∴ P_final = [ {(10000 kg m/s) cosθ} i + {(10000 kg m/s) sinθ} j ] + [ {(3000 v) cos(26.6)} i - {(3000 v) sin(26.6)} j ]

∴ P_final = { (10000 kg m/s) cosθ + (3000 v) cos(26.6) } i + { (10000 kg m/s) sinθ - (3000 v) sin(26.6)} j

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for elastic collision : P_initial = P_final

Thus, Comparing i and j components of initial and final momentum. We get :

∴ { (10000 kg m/s) sinθ - (3000 v) sin(26.6)} = 0

∴ sinθ = (0.3) v sin(26.6)

∴ sinθ = (0.1343) v

∴ sinθ = (0.1343)(4.472)

∴ sinθ = 0.6007

∴ θ = 36.9^o