Help calculating impulse

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  • #1
noname1
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A 0.300 kg baseball just before and after it collides with a bat. Just before, the ball has a velocity v1 of magnitude 12.0 m/s and angle θ1 = 35°. Just after, it is traveling directly upward with velocity v2 of magnitude 10.0 m/s. The duration of the collision is 1.5 ms.
(a) What is the magnitude of the impulse on the ball from the bat?

I tried solving a by

pi = (.3)(12)(cos215i+sin215j) = -2.95i + 2.06j
pf = (.3)(12)j = 3.6j
i = pf - pi = 3j - (-2.95i - 2.06j) = 2.95i +5.66j

mag = sqrt(2.95²+5.66²) = 6.38

what am i doing wrong?
 

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Answers and Replies

  • #2
berkeman
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A 0.300 kg baseball just before and after it collides with a bat. Just before, the ball has a velocity v1 of magnitude 12.0 m/s and angle θ1 = 35°. Just after, it is traveling directly upward with velocity v2 of magnitude 10.0 m/s. The duration of the collision is 1.5 ms.
(a) What is the magnitude of the impulse on the ball from the bat?

I tried solving a by

pi = (.3)(12)(cos215i+sin215j) = -2.95i + 2.06j
pf = (.3)(12)j = 3.6j
i = pf - pi = 3j - (-2.95i - 2.06j) = 2.95i +5.66j

mag = sqrt(2.95²+5.66²) = 6.38

what am i doing wrong?

Would the delta-time be part of the relevant equations....?
 
  • #3
noname1
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pf should be 10 and not 12 right? shoot, how could i miss that, just want to be sure because i just have one more attempt

pi = (.3)(12)(cos215i+sin215j) = -2.95i + 2.06j
pf = (.3)(10)j = 3j
i = pf - pi = 3j - (-2.95i - 2.06j) = 2.95i +5.06j

mag = sqrt(2.95²+5.06²) = 5.86


correct?
 
  • #4
berkeman
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62,833
13,710
pf should be 10 and not 12 right? shoot, how could i miss that, just want to be sure because i just have one more attempt

pi = (.3)(12)(cos215i+sin215j) = -2.95i + 2.06j
pf = (.3)(10)j = 3j
i = pf - pi = 3j - (-2.95i - 2.06j) = 2.95i +5.06j

mag = sqrt(2.95²+5.06²) = 5.86


correct?

You didn't answer my question. Where should the 1.5ms enter into your calculation of the impulse?
 
  • #5
noname1
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i dont think it shouldnt enter because

p = mass vs velocity

pf = 0.3 x 10 = 3j
pi = .3 x 12(cos215i+sin215j) = -2.95i + 2.06j

right?
 
  • #6
berkeman
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62,833
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i dont think it shouldnt enter because

p = mass vs velocity

pf = 0.3 x 10 = 3j
pi = .3 x 12(cos215i+sin215j) = -2.95i + 2.06j

right?

My apologies. I was misremembering the definition of Impulse.

It looks like you just have a sign error in the y component of the incoming ball's momentum:

pi = (.3)(12)(cos215i+sin215j) = -2.95i + 2.06j

The diagram in your original post (OP) shows the ball coming in at a downward angle.
 
  • #7
noname1
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yes you are correct but anyways its not going to affect the answer since we take the absolute value of it, but something and i think this is correct

pi = (.3)(12)(cos215i+sin215j) = -2.95i + 2.06j
pf = (.3)(10)j = 3j
i = pf - pi = 3j - (-2.95i - 2.06j) = 2.95i +5.06j

mag = sqrt(2.95²+5.06²) = 5.86


but just want to verify since its my last attempt to the question
 
  • #8
berkeman
Mentor
62,833
13,710
yes you are correct but anyways its not going to affect the answer since we take the absolute value of it, but something and i think this is correct

pi = (.3)(12)(cos215i+sin215j) = -2.95i + 2.06j
pf = (.3)(10)j = 3j
i = pf - pi = 3j - (-2.95i - 2.06j) = 2.95i +5.06j

mag = sqrt(2.95²+5.06²) = 5.86


but just want to verify since its my last attempt to the question

You do not take any absolute values or do sqrt(squares) magnitude calculation until after you have subtracted the vectors. So the direction of the y component of the incoming baseball (negative) is important, since the final direction is upward (positive).
 

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