A 7.50 kg bowling ball strikes a stationary 4.50 kg bowling ball. After the collision, the 4.50 kg

ball is traveling 8.50 m/s 30.0° east of north; the 7.50 kg ball is traveling 6.50 m/s 20.0° south of
east. What was the initial velocity (magnitude and direction) of the 7.50 kg ball?

North south problem

7.5 Vn = 4.5 (8.5 cos 30) -7.5(6.5 sin 20)

East west problem
7.5 Ve = 4.5(8.5 sin 30) +7.5(6.5 cos 20)

To find the initial velocity of the 7.50 kg bowling ball, we can use the principle of conservation of momentum.

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

The momentum of an object is given by the product of its mass and velocity. The momentum of the 7.50 kg ball before the collision can be represented as:
p1 = m1 * v1

The momentum of the 4.50 kg ball before the collision is zero because it is initially stationary:
p2 = 0

After the collision, the 4.50 kg ball is traveling 8.50 m/s 30.0° east of north, and the 7.50 kg ball is traveling 6.50 m/s 20.0° south of east.

To simplify the calculation, we can break down the velocities into their horizontal and vertical components.

Component of the 4.50 kg ball's velocity in the vertical direction:
v2y = 8.50 m/s * sin(30.0°)

Component of the 7.50 kg ball's velocity in the horizontal direction:
v1x = 6.50 m/s * cos(20.0°)

Component of the 7.50 kg ball's velocity in the vertical direction:
v1y = -6.50 m/s * sin(20.0°)

To find the magnitude and direction of the initial velocity of the 7.50 kg ball, we can use the Pythagorean theorem and trigonometry.

Magnitude of the initial velocity:
v1 = √(v1x^2 + v1y^2)

Direction of the initial velocity:
θ = arctan(v1y / v1x)

By plugging in the values and solving for v1 and θ, we can find the initial velocity of the 7.50 kg ball.