The boy jumps off the flat car at A with a velocity of v=4...

The boy jumps off the flat car at $A$ with a velocity of $v=4$ ft $/$ s relative to the car as shown. If he lands on the second flat car $B$, determine the final speed of both cars after the motion. Each car has a weight of 80 lb. The boy's weight is 60 lb, Both cars are originally at rest. Neglect the mass of the car's wheels.

Key Concepts

Relative Velocity and Reference Frames

When velocities are given relative to one object (in this case, the boy’s speed relative to the car), it is important to transform these velocities to an appropriate inertial frame of reference. This involves adding or subtracting the velocity of the moving frame (the car) in order to obtain the absolute velocities of the individual bodies. This concept is essential for properly applying conservation laws and analyzing the motion of all objects involved.

Conservation of Momentum

This principle states that when no external forces act on a closed system, the total momentum of the system remains constant. In problems involving multiple bodies interacting through internal forces (like the boy jumping from one car to another), the sum of the momenta before the interaction is equal to the sum after the interaction. This is crucial for solving problems where individual components gain velocities relative to an inertial frame, such as a moving platform or vehicle.

Transcript

00:01 For this problem, let's first consider the conservation of momentum and apply it to the boy and to cart a.

00:12 So the conservation of linear momentum, to be specific, tells us that the momentum of the initial state of the system, mv1, must equal to mv2.

00:36 So initially the boy and the car were both addressed.

00:43 So they have no velocity and hence no momentum.

00:48 And then we have the momentum of the cart which is to the right.

00:54 So we'll take the left direction in this problem to always be positive.

01:00 So for the car a, it has a mass of 80 pounds divided by g, which is 32 .2 feet per square second.

01:09 Times its velocity with which it moves off va plus the momentum due to the boy, which is his weight, 60 pounds over g, which is 32 .2 feet per square second, times his velocity, which is v little b, and this is in the x direction.

01:30 Remember, his velocity is x and y components, whereas the cars have only x components.

01:36 So we're just looking at the conservation of linear momentum, in the x direction so now if we rearrange this equation we show that the velocity of car a is equal to 0 .75 times the x component of the boy boy's velocity vb now we also know that the velocity of the boy is final velocity vb is equal to v a plus vb, b of the boy relative to cart a.

02:25 So the boy's velocity is equal to the velocity of the cart plus his velocity relative to cart a.

02:31 So if we take the x components of this, vb in the x direction is equal to minus v a that only has x components, plus 4 times cosine of theta which is cosine of theta in this case is 12 over 13 in our diagram that's 4 times 12 by 13 and so hence we get vb in the x direction and we get v a from above in terms of vb so we get vb in the x direction to be 2 .11 feet per second...

Please give Ace some feedback