A ball with a mass of 0.600kg is initially at rest. It is struck by a second bal

Question

A ball with a mass of 0.600kg is initially at rest. It is struck by a second ball having a mass of 0.400kg , initially moving with a velocity of 0.250m/s toward the right along the x axis. After the collision, the 0.400kg ball has a velocity of 0.200m/s at an angle of 36.9? above the x axis in the first quadrant. Both balls move on a frictionless, horizontal surface.

What is the change in the total kinetic energy of the two balls as a result of the collision? Answer in J

A catcher catches a 145g baseball traveling horizontally at 34.0m/s .

If it takes the ball 25.0ms to stop once it is in contact with the catcher's glove, what average force did the ball exert on the catcher? Answer in N

Explanation / Answer

momentum is conserved
momentum is a vector quantity
so
we draw (0.400*0.25)@ 0 deg and (0.400*0.200)@ 36.9 deg
then connect the ends and draw the parallel to that from the origin
where it crosses the vector on the x axis ( the sum ) marks the end of this other vector
now use trig to solve for length and angle
this is a vector subtraction
(0.400*0.250)@0 - (0.400*0.200)@36.9 = 0.0600@-53.13 deg
so ( m v ) = 0.0600 and m = 0.600 so v = 0.100 m/s@-53.13 deg

C
energy is not a vector sum - it is strictly algebraic
so
initial E - final E = delta E
1/2 (0.400) (0.250)^2 - [ (1/2 (0.400) (0.200)^2) + 1/2 (0.600) (0.100)^2 ] = 0.0015 J lost

2. The impulse I = F x t and F = I / t =

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.