Question

Two 2.3 kg bodies, A and B, collide. The velocities before the collision are and . After the collision, . What are (a) the x-component and (b) the y-component of the final velocity of B? (c) What is the change in the total kinetic energy (including sign)?

Answer 1

Missing details in the question:

The velocities before collision are

$u_A=(40i+49j) m/s$

$u_B=(35i+11j) m/s$

After the collision:

$v_A=(14i+21j) m/s$

(a) 61 m/s

We can solve the problem by simply treating separately the x- and the y-components of the motion.

Here we want to analzye the motion along x. We have:

$u_A = 40 m/s$ is the initial velocity of A along the x-direction

$u_B = 35 m/s$ is the initial velocity of B along the x-direction

$v_A = 14 m/s$ is the final velocity of A along the x-direction

$v_B = ?$ is the final velocity of B along the x-direction

Since the total momentum along the x-direction must be conserved, we can write

$mu_A + mu_B = mv_A + mv_B$

where

m = 2.3 kg is the mass of the two bodies. Since the mass is the same, we can eliminate it from the equation,

$u_A + u_B = v_A + v_B$

And so, we find the final velocity of B along the x-direction:

$v_B = u_A + u_B - v_A=40+35-14=61 m/s$

(b) 39 m/s

Similarly to what we did in part a), here we analyze the conservation of momentum along the y-direction.

We have:

$u_A = 49 m/s$ is the initial velocity of A along the y-direction

$u_B = 11 m/s$ is the initial velocity of B along the y-direction

$v_A = 21 m/s$ is the final velocity of A along the y-direction

$v_B = ?$ is the final velocity of B along the y-direction

Since the total momentum along the y-direction must be conserved, we can write

$mu_A + mu_B = mv_A + mv_B$

Since the mass is the same, we can eliminate it from the equation,

$u_A + u_B = v_A + v_B$

And so, we find the final velocity of B along the y-direction:

$v_B = u_A + u_B - v_A=49+11-21=39 m/s$

c) +615 J

Here we have to find the total kinetic energy before and after the collision first.

First, we have to find the speed of each object before and after the collision. We have:

$u_A = \sqrt{40^2+49^2}=63.2 m/s\\u_B = \sqrt{35^2+11^2}=36.7 m/s\\v_A = \sqrt{14^2+21^2}=25.2 m/s\\v_B = \sqrt{61^2+39^2}=72.4 m/s$

So, the total kinetic energy before the collision was

$K_i = \frac{1}{2}mu_A^2+\frac{1}{2}mu_B^2 = \frac{1}{2}(2.3)(63.2)^2+\frac{1}{2}(2.3)(36.7)^2=6143 J$

While after the collision

$K_f = \frac{1}{2}mv_A^2+\frac{1}{2}mv_B^2 = \frac{1}{2}(2.3)(25.2)^2+\frac{1}{2}(2.3)(72.4)^2=6758 J$

So, the change in kinetic energy is

$\Delta K = K_f - K_i = 6758-6143 = +615 J$

(note that the system cannot gain kinetic energy in the collision, unless there is an external force acting on it)