Question

A billiard ball moving at 5.60 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 5.03 m/s at an angle of 26.0° with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity after the collision. Magnitude m/s direction ° (with respect to the original line of motion)

Answer 1

In an elastic collision between two billiard balls, both momentum and kinetic energy are conserved. Given the initial and final velocities, we can break down the velocity into horizontal and vertical components. By subtracting the first ball's velocity from the total velocity, we can determine the velocity of the struck ball after the collision.

Step-by-step explanation:

In an elastic collision, both momentum and kinetic energy are conserved. To find the velocity of the struck ball after the collision, we can use the principle of conservation of momentum.

Given that the initial velocity of the first ball is 5.60 m/s and the final velocity is 5.03 m/s at an angle of 26.0° with respect to the original line of motion, we can break down the velocity into horizontal and vertical components.

The horizontal component of the velocity is given by 5.03 m/s * cos(26.0°) = 4.58 m/s. The vertical component is given by 5.03 m/s * sin(26.0°) = 2.12 m/s.

Since the collision is elastic, the total momentum before and after the collision should be the same. Therefore, the struck ball's velocity after the collision can be calculated by subtracting the first ball's velocity from the total velocity. The horizontal component of the struck ball's velocity is 4.58 m/s - 5.60 m/s = -1.02 m/s. The vertical component is 2.12 m/s. Therefore, the struck ball's velocity after the collision is approximately -1.02 m/s Î + 2.12 m/s Ĵ.