ENERGY and MOMENTUM: What is the Final Velocity of the Ball?

Q: A 10 kg iron ball moves in an Eastward direction at 5.0 m/s. It collides with a 5.0 kg rubber ball moving Northward at 10 m/s. After the collision, the iron ball moves 60° East of North with a speed of 4.0 m/s. What is the velocity of the rubber ball after the collision?

A: We need to have the ( x ) and ( y ) components of momentum for the initial system. Additionally, we use the data given to obtain the ( x ) and ( y ) components of momentum of the iron ball after the collision. Afterwards, we can subtract the ( x ) and ( y ) momentum components of the iron ball from the ( x ) and ( y ) components of the initial system. We should then have the ( x ) and ( y ) components of momentum associated with the rubber ball. Subsequently, we use the Pythagorean theorem to obtain a net momentum vector for the rubber ball. By dividing the final momentum of the rubber ball by its mass, we determine its final speed.

The initial momentum ( p_1i ) in the +x-direction ( Eastward ) is ( 10 kg )( 5 m/s ) = ( 50 kg*m/s ). The initial momentum ( p_2i ) in the +y-direction ( Northward ) is ( 5 kg )( 10 m/s ) = ( 50 kg*m/s ). The final momentum ( p_1f ) of the iron ball is ( 10 kg )( 4.0 m/s ) at an angle of 60 degrees East of North. We use trigonometry to determine the final ( x ) and ( y ) components of the rubber ball:

sin 60 degrees = ( x_1f / p_1f ), and cos 60 degrees = ( y_1f / p_1f )

( 40 kg*m/s )( sin 60 degrees ) = ( x_f1 ) = ( 35 kg*m/s )

( 40 kg*m/s )( cos 60 degrees ) = ( y_1f ) = ( 20 kg*m/s )

Let’s now subtract the ( x_1f ) and ( y_1f ) components of the iron ball’s momentum from the total initial ( x ) and ( y ) components of momentum:

( 50 kg*m/s ) – ( 35 kg*m/s ) = ( 15 kg*m/s ) in the x-direction.

( 50 kg*m/s ) – ( 20kg*m/s ) = ( 30 kg*m/s ) in the y-direction.

The momentum values that remain represent the final x and y components of the rubber ball. Using the Pythagorean theorem, we obtain the NET momentum vector of the rubber ball:

( p_2f ) = √ [ ( 15kg*m/s )² + ( 30 kg*m/s )² ) ] 

( p_2f ) = ( 34 kg*m/s )

Since ( p_2f ) = ( 5 kg )( ? m/s ), divide both sides of the equation by ( 5 kg ) to obtain the final velocity of the rubber ball:

( v_2f ) = [ ( 34 kg*m/s ) / ( 5 kg ) ] = 6.8 m/s.