A 4154 kg truck traveling with a velocity of 12ms due east...

A 4154-kg truck traveling with a velocity of 12(m)/(s) due east collides head-on with a 904-kg car traveling with a velocity of 40(m)/(s) due west. The two vehicles stick together after the collision. What is their shared velocity in (m)/(s) after the collision? Define East to be the positive direction. Type your answer without the units. ◻ QUESTION 14 The change in momentum for the driver of a car hitting a brick wall at 10(m)/(s) is the same with and without air bags. Suppose a driver comes to a complete stop in 0.1s without an air bag. Suppose also that the same driver comes to a complete stop in 0.5s with an airbag. What is the ratio of the average force during the deceleration with and without an airbag? QUES11ON13 A 4154-kg truck traveling with a velocity of 12 m/s due east collides head-on with a 904-kg car traveling with a velocity of 40 m/s due west. The two vehicles stick together after the collision. What is their shared velocity in m/s after the collision? Define East to be the positive direction. Type your answer without the units QUESTION14k The change in momentum for the driver of a car hitting a brick wall at 10 m/s is the same with and without air bags. Suppose a driver comes to a complete stop in 0.1 s without an air bag.Suppose also that the same driver comes to a complete stop in 0.5 s with an airbag. What is the ratio of the average force during the deceleration with and without an airbag?

Transcript

00:01 This problem, the mass of the truck is five times the mass of the car.

00:05 So the truck of mass 5m is moving with a velocity u1 equal to 20 meter per second towards east in this direction.

00:18 It collides with a car of mass m because the mass of the car is 1 by 5th of the mass of the truck.

00:26 It is moving with a velocity u2.

00:29 This we need to find out.

00:31 After collusion, they stick with each another and they become one body of mass 5m plus m equal to 6m and they move with a combined velocity v.

00:45 Now after they collide, they further move by a distance of 22 .5 meter towards east before they come to a storm.

01:02 It means this is just before the collision is this position and now when they have moved a distance of 22 .5 meters, both of them, then they finally stopped.

01:25 So we need to find out their velocity after collision and also the velocity of the car before the collision right so for part a we need to find out first we need to find out the acceleration after the collision it is the force of friction between the car plus truck force of kinetic friction between the car plus truck causes it to stop right so it means the breaking force with cause it is to stop the breaking force is equal to the force of kinetic friction force of kinetic friction this is the force which actually stops them after they collide so what will be the breaking acceleration will be equal to the breaking force upon mass and the breaking force is equal to force of kinetic friction and force of kinetic friction we know that coefficient of kinetic friction into normal reaction upon m so and in this case we know that the weight is acting downward m g and the normal reaction is acting upward same is here their weight i mean 6mg is acting downward and normal reaction is acting upward.

03:15 So n equal to m g.

03:17 So this equal to.

03:29 So what is the means breaking acceleration? the breaking acceleration is equal to coefficient of kinetic friction n equal to mg upon m it means it is coefficient of kinetic friction into g and in the problem coefficient of kinetic friction is given as as 0 .5 and g should be taken as 10.

03:57 So the acceleration with sign is minus 5 meter per second square.

04:03 The negative sign indicate it is the braking acceleration.

04:07 So this is the acceleration of the truck plus car after the collision which causes them to stop.

04:20 Now with this acceleration, we need to apply the equation of my equation of my motion v square minus u square equal to 2 as on the motion of the truck plus car immediately after the collision and up to the time when they stop right so immediately after the collision let their velocity is u and when they stop their final velocity is v a is the braking acceleration which we have already calculated and as is the distance that travel say before coming to halt is 22 .5 meter so we we use here final velocity is zero because they come to the halt and you we need to find out it is the their combined velocity after the impact so this will be zero square minus u square is equal to 2 a is equal to minus 5 and s is 22 .5 meter.

05:34 So the u comes out to be 15 meter per second...

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