1. A billiard ball of mass mA = 0.4 kg moving with a speed vA = 2 m/s strikes a second ball, initially at rest, of mass mB = 0.4 kg. As a result of the collision, the first ball is deflected off at an angle of 30.0o with a speed v’A = 1.2 m/s. Taking the xaxis to be the original direction of motion of ball A, and assuming it deflects above the xaxis, find the final velocity of ball B (which will include both a magnitude as well as a direction).

Question

1. A billiard ball of mass mA = 0.4 kg moving with a speed vA = 2 m/s strikes a second ball, initially at rest, of mass mB = 0.4 kg. As a result of the collision, the first ball is deflected off at an angle of 30.0o with a speed v’A = 1.2 m/s. Taking the xaxis to be the original direction of motion of ball A, and assuming it deflects above the xaxis, find the final velocity of ball B (which will include both a magnitude as well as a direction).

Answer 1

1. A billiard ball of mass mA = 0.4 kg moving with a speed vA = 2 m/s strikes a second ball, initially at rest, of mass mB = 0.4 kg. As a result of the collision, the first ball is deflected off at an angle of 30.0o with a speed v’A = 1.2 m/s. Taking the xaxis to be the original direction of motion of ball A, and assuming it deflects above the xaxis, find the final velocity of ball B (which will include both a magnitude as well as a direction).

Answer 2

To find the final velocity of ball B, we can use the principle of conservation of momentum.

First, we need to find the momentum before and after the collision. The momentum of an object is given by the product of its mass and velocity.

Before the collision:
Momentum of ball A = mA * vA
Momentum of ball B = mB * 0 (since ball B is initially at rest)

After the collision:
Momentum of ball A' = mA * v'A
Momentum of ball B' = mB * vB' (final velocity of ball B)

According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.

mA * vA = mA * v'A + mB * vB'

Substituting the given values:
0.4 kg * 2 m/s = 0.4 kg * 1.2 m/s + 0.4 kg * vB'

Simplifying the equation:
0.8 kg m/s = 0.48 kg m/s + 0.4 kg * vB'

Subtracting 0.48 kg m/s from both sides:
0.32 kg m/s = 0.4 kg * vB'

Dividing both sides by 0.4 kg:
0.8 m/s = vB'

So, the final velocity of ball B is 0.8 m/s, in the same direction as ball A deflects (above the x-axis).

Answer 3

To find the final velocity of ball B, we can use the principle of conservation of momentum and apply it to the collision between the two balls.

The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces act on the system.

Mathematically, we can express this as:

mAvA + mBvB = mAv'A' + mBv'B'

Where:
mA and mB are the masses of ball A and ball B, respectively.
vA and vB are the initial velocities of ball A and ball B, respectively.
v'A' and v'B' are the final velocities of ball A and ball B, respectively.

Since ball B is initially at rest (vB = 0) and only ball A is deflected off, the equation simplifies to:

mA * vA = mAv'A' + mBv'B'

Given:
mA = 0.4 kg (mass of ball A)
vA = 2 m/s (initial velocity of ball A)
v'A' = 1.2 m/s (final velocity of ball A)
mB = 0.4 kg (mass of ball B)

Substituting these values into the equation, we get:

(0.4 kg) * (2 m/s) = (0.4 kg) * (1.2 m/s) + (0.4 kg) * v'B'

Simplifying further:

0.8 kg·m/s = 0.48 kg·m/s + (0.4 kg) * v'B'

Rearranging the equation to solve for v'B':

(0.4 kg) * v'B' = 0.8 kg·m/s - 0.48 kg·m/s

(0.4 kg) * v'B' = 0.32 kg·m/s

v'B' = (0.32 kg·m/s) / (0.4 kg)

v'B' = 0.8 m/s

Therefore, the final velocity of ball B is 0.8 m/s, in the same direction as the initial velocity of ball A.