A 0.29 kg softball has a velocity of 10 m/s at an angle of 25° below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while it is in contact with the bat if the ball leaves the bat with ?

1 Answer
Oct 14, 2015

(a) #4.31 kg m s^(–1_#
(b) #7.37 kg m s^(–1_#

Explanation:

Solve the problem by using vector diagrams.

  1. Draw vector for initial momentum.
    enter image source here

  2. To calculate the change in momentum use:
    #∆p = p_f – p_i = p_f + (–p_i)#. For that reason also draw the negative vector for initial momentum, i.e. draw #–p_i#.
    [#p_i# is the initial momentum, #p_f# is the final momentum.]
    enter image source here

  3. Use vector addition to solve problems (a) and (b). First draw the vector for final momentum. Second, draw the vector triangle. Third, use the cosine rule to solve for ∆p #(a^2 = b^2 + c^2 – 2bc cosA)#.

Solution for (a)
enter image source here
enter image source here
Cosine rule:
#∆p^2 = 4.64^2 + 2.9^2 – 2 × 4.64 × 2.9 × cos(65º) = 29.939… – 11.373… = 18.566…#
[Don’t forget to square-root the value above as it is not the solution to the problem yet!]
#⇒ ∆p = √18.566 = 4.31 kg. m .s^(–1)#

Solution for (b)
enter image source here
enter image source here
Cosine rule:
#∆p^2 = 4.64^2 + 2.9^2 – 2 × 4.64 × 2.9 × cos(155º) = 29.939… – (–24.390…) = 54.330…#
#⇒ ∆p = √54.330 = 7.37 kg . m .s^(–1)#