1. A small car of mass 645 kg is parked behind a small truck of mass 1754 kg on
ID: 1320594 • Letter: 1
Question
1. A small car of mass 645 kg is parked behind a small truck of mass 1754 kg on a level road. The brakes of both the car and the truck are off so that they are free to roll with negligible friction. A 88 kg woman sitting on the tailgate of the truck shoves the car away by exerting a constant force on the car with her feet. The car accelerates at 1.2 m/s 2 .
a)What is the acceleration of the truck? Answer in units of m/s 2
b)What is the magnitude of the force exerted on the car? Answer in units of N
2. A 2.70 kg block is in equilibrium on an incline of 34.8? . The acceleration of gravity is 9.81 m/s 2 . What is Fn of the incline on the block? Answer in units of N
Explanation / Answer
This is an example. It may help you.
a. According to law ofconservation of linear momentum
momentum of truck andwoman + momentum ofcar = 0
( as initially car and truck were atrest)
(mt + mw)* vt + mc *vc = 0
m represent mass and v represent velocity.Subscript t, w and c refer to truck, woman and carrespectively.
=> (1592 + 35) *vt + 826 *vc = 0
But vt = at*t and vc = ac* t
where a is respective acceleration.
1627 * at *t = - 826 * 1.6 * t
acceleration of thetruck at = -1321.6 / 1627
= -0.812 m/s2
-ve sign is due to the fact that the truckis accelerated in direction opposite to the car.
b. Force oncar Fc = mc* ac
= 826* 1.6
= 1321.6 N
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