Question

Ball B, moving in the positive direction of an x axis at speed v, collides with stationary ball A at the origin. A and B have different masses. After the collision, B moves in the negative direction of the y axis at speed v/5. In what direction does A move, as an angle with respect to the x axis?

Answer 1

Conservation of momentum in the x-direction:

m_Bv + 0 = m_Av_Ax + 0

m_Av_Ax = m_Bv     Eqn 1

In the y-direction:

0 + 0 = m_Av_Ay - m_B(v/5)

m_Av_Ay = m_Bv/5     Eqn 2

Divide Eqn 2 by Eqn 1, and you'll get:

v_Ay/v_Ax = 1/5 = 0.2

The angle of the final velocity vector is found from tan θ = v_y/v_x

θ = tan^-10.2 = 11.3 degrees

Answer 2

yolo

Answer 3

Momentum is conserved during collisions. It is also a vector. I'll be using the notation (x, y) for vectors.

Ball B has mass = m1 and initial momentum P_B

Ball A has mass = m2 and initial momentum P_A

First calculate the total initial momentum:

1) P_initial = P_B + P_A

P_B = (m1*v, 0)

P_A = (0, 0)

So:

2) P_initial = (m1*v, 0)Now calculate the final momentum, P_final. We know the initial momentum and the final momentum of Ball B.

Ball B now has final momentum P_BF.

Ball A has final momentum P_AF.

Knowing that P_final = P_initial, eqn. 1) can be rearranged to solve for the final momentum of Ball A:

3) P_AF = P_initial - P_BF

4) P_BF = (0, m1*-(1/5)v) as given in the problem

Now combine eqns. 2), 3), and 4) to solve for the momentum vector for Ball A, P_AF.

Recall that vectors add like so: (a, b) + (c, d) = (a+c, b+d)

There is going to be an x and y component to this vector. Using Newton's Third Law you may be able to write this vector without any vector operations, but if not, you may use the above equations to determine it.Now to calculate the angle between Ball A's momentum vector, P_AF, and the x-axis.

You have calculated a vector for P_AF in the form (x1, y1) (these may be positive or negative values). It may help to draw out the vectors on a 2-d graph. Draw the x1 component along the x-axis, then starting the at the tip of this x1 vector component, draw the y1 component. These will be the sides of the triangle used to calculate the angle to the x-axis in the equation below.

Using this equation:

So:

θ = tan^-1(y1/x1)