physics question

Question

A steel ball of mass 10 kg moves due East at 5.0 m/s. It collides with a rubber ball of mass 5.0 kg moving at 10 m/s due North. After the collision the steel ball moves at an angle of 60° East of North with a speed of 4.0 m/s. What is the velocity of the rubber ball after the collision? (answer is 6.74 but i keep getting 4.714)

Answer 1

We need to have the ( x ) and ( y ) components of momentum for the initial system. Additionally, we should use the data given to get the ( x ) and ( y ) components of the momentum of the steel ball after the collision. Afterward. we can subtract the ( x ) and ( y ) momentum components of the steel ball from the ( x ) and ( y ) components of the initial system. We should then have the ( x ) and ( y ) components of motion associated with the rubber ball. By using the Pythagorean theorem, we can now get a net ( x, y ) vector for the rubber ball. By dividing this value by the mass of the rubber ball, we should be able to determine the final speed of the rubber ball.

The initial momentum ( p1i ) in the ( x initial ) direction is ( 10 kg )( 5 m / s ) = ( 50 kg*m/s ). The initial momentum ( p2i ) in the ( y initial ) direction is ( 5 kg )( 10 m / s ) = ( 50 kg*m/s ). The final momentum of the steel ball is p1f = ( 10 kg )( 4 m / s ) at an angle of 60 degrees East of North. We use trigonometry to determine the final ( x ) and ( y ) components of the steel ball.

Sin 60 degrees = ( x / p1f ), and Cos 60 degrees = ( y / p1f ).

( 40 kg*m/s )( sin 60 degrees ) = ( x final ) = ( 35 kg*m/s ).

( 40 kg*m/s )( cos 60 degrees ) = ( y final ) = ( 20 kg*m/s ).

Let's now subtract these ( x final ) and ( y final ) components of the steel 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.

These values should represent the NET components of the remaining rubber ball. Let's use them with the Pythagorean theorem to obtain a NET resultant vector that lies in the ( x, y ) plane.

( p final rubber ball ) = sqrt [ ( 15kg*m/s )^2 + ( 30 kg*m/s )^2 ) ]. ( p final rubber ball ) = ( 34 kg*m/s ).

Since ( p final rubber ball ) = ( 5 kg )( ? m / s ), let's divide both sides of the equation by ( 5 kg ) to see what velocity we derive.

( v final rubber ball ) = [ ( 34 kg*m/s ) / ( 5 kg ) ] = 6.8 m / s.

WELL, I MUST HAVE SOME ROUNDING ERROR SOMEWHERE :) . THIS ANSWER, HOWEVER, IS VERY CLOSE TO THE ONE YOU HAVE ABOVE ( 6.74 m / s ).

Best of luck with your studies!

P.S. I answered another question of yours about an hour ago, but it was erased somehow or another. In your last problem, there was a model airplane moving diagonally downward in the East of South direction ( I think ); After calculating the momentum for this plane, I broke this airplane's momentum into ( x1i ) and ( y1i ) components.

The other plane was moving straight Southward, and thus, its momentum had no ( x ) component and was labeled ( y2i ). Afterward, I added the ( y1i ) component of plane ( 1 ) to the Southbound ( y2i ) momentum of plane ( 2 ). I then used the Pythagorean Theorem to add ( x1i ) to ( y1i + y2i ).

( pf total ) = sqrt [ ( x total^2 ) + ( yf )^2 ].

I subsequently set ( pf total ) = [ ( m1 + m2 )( ? km / h ) ].

After dividing both sides by ( m1 + m2 ), I got the correct velocity value for ( km final ).

After this, I used the tan ( theta ) = ( ( x ) / ( y total ) ) to get ( theta ).

Best of luck with your studies!!!