A blue car with mass m c = 484 kg is moving east with a speed of v c = 17 m/s an
ID: 1343182 • Letter: A
Question
A blue car with mass mc = 484 kg is moving east with a speed of vc = 17 m/s and collides with a purple truck with mass mt = 1330 kg that is moving south with an unknown speed. The two collide and lock together after the collision moving at an angle of = 52° South of East
1) What is the magnitude of the initial momentum of the car?
2) What is the magnitude of the initial momentum of the truck?
3) What is the speed of the truck before the collision?
4) What is the magnitude of the momentum of the car-truck combination immediately after the collision?
5) What is the speed of the car-truck combination immediately after the collision?
Explanation / Answer
Given mc=484 kg
mt=1330 kg
Initial speed of the car vc = 17 m/s
Angle of the car and truck after collision theta = 52 o
In east direction :
Apply law of conservation of linear momentum ,
mc x vc +(mt x0) = ( mc+mt) V cos 52 Where V = Speed of car and truck combination after collision.
8228 + 0 = 1814 x 0.61566 V
8228 = 1116.8 V
From this V = 7.367m/s
In south direction :
Apply law of conservation of linear momentum ,
(mc x 0) +(mt x vt) = ( mc+mt) V sin 52 Where V = Speed of car and truck combination after collision.
0 + 1330 vt = 1814 x 7.3674x sin52
= 10531.35
vt = 7.918 m/s
1) The magnitude of the initial momentum of the car P = mc x vc
= 8228 kg m/s
2) The magnitude of the initial momentum of the truck P ' = mt x vt
= 10531.35 kg m/s
3)The speed of the truck before the collision vt = 7.918 m/s
4)The magnitude of the momentum of the car-truck combination immediately after the collision = P + P '
= 18759.35 kgm/s
5)The speed of the car-truck combination immediately after the collision V = 7.367m/s
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