A car of mass m moving at a speed v1 collides and couples...

A car of mass $m$ moving at a speed $v_{1}$ collides and couples with the back of a truck of mass 2$m$ moving initially in the same direction as the car at a lower speed $v_{2},(\text { a })$ What is the speed $v_{f}$ of the two vehicles immediately after the collision? (b) What is the change in kinetic energy of the car-truck system in the collision?

Key Concepts

Conservation of Linear Momentum

This principle states that in the absence of external forces, the total momentum of a system remains constant. In collision problems, like when two vehicles lock together, conservation of momentum is applied to relate the combined mass and velocity after the collision to the individual momenta before impact.

Inelastic Collision

An inelastic collision is a type of collision in which the colliding objects stick together after impact. While momentum is conserved, kinetic energy is not; some is transformed into other forms such as heat or sound. This concept is essential when calculating the final velocity of coupled bodies and determining the associated kinetic energy loss.

Kinetic Energy Loss in Collisions

In inelastic collisions, the system's kinetic energy decreases because some of it is dissipated due to deformation, friction, and other non-conservative forces. The change in kinetic energy is determined by comparing the total kinetic energy before the collision with that after the collision, highlighting the efficiency—or lack thereof—in energy transfer during impact.

Transcript

00:01 We know for part a, even if it's a perfectly inelastic collision, we can still apply the moment of conservation of momentum.

00:08 And we can say that, of course, momentum final equals momentum initial.

00:12 And this will become m plus 2m multiplied by v final, equaling mv initial or mv sub 1 plus 2m v sub 2.

00:26 Of course, m's rather, let's not cancel this out yet.

00:30 We can simply say that v final is equaling one -third of v -sub -1 plus 2 v -sub 2.

00:42 Now this would be our answer for part i.

00:46 Now for part b we want to find the change in the kinetic energy.

00:50 So for part b the energy before the impact, so kinetic energy initial is going to be equaling 1 1� m v .1 squared plus 1 half times 2m v.

01:05 2 squared.

01:06 So this would equal m over 2 multiplied by v .1 squared plus 2 v.

01:15 Sub 2 squared.

01:17 Keep that in mind and we have the kinetic energy final.

01:21 So this would be after the collision.

01:23 This would simply be 1 3 3 .3 times 3.

01:27 M v final squared.

01:31 This would be equaling 3m over 2 multiplied by 1 over 9 v sub 1 squared plus 4 v sub 1 v sub 2 plus 4 v sub 2 squared.

01:55 So at this point we can simplify a bit and say that 4 part b, again kinetic and energy final will then be equaling m over 2 and then this would be v sub 1 squared over 3 plus 4 v sub 1 v sub 2 divided by 3 plus 4 v sub 2 squared divided by 3.

02:27 Now we can then say that the change in the kinetic energy is going to be equalling the kinetic energy final minus the kinetic energy initial.

02:36 So before the collision minus after the, rather after the collision minus before the collision...

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