Physics, 26.10.2019 02:43 chloeann5397

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)?

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Answer from: meganwintergirl

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)

Answer from: Quest

Earthplanet yplanet y5=earth *many different equationsplanet x1.5=earthplanet x is two times planet yobject 5,2500 mass5,1,3,2,4object 1 is four times obect 4

Answer from: Quest

molecules are free to separate from each other and will take on the shape of the container. molecules move completely at random and high speeds. they vibrate in a fixed position and do not move around much. molecules expand to the edge of their container.

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Two 2.3 kg bodies, a and b, collide. the velocities before the collision are and . after the collisi...