Momentum (Catching Ball on Ice)

AI Thread Summary
The discussion revolves around solving a physics problem involving momentum on ice. In Part A, the correct speed after catching a ball is calculated using the conservation of momentum. For Part B, the user initially misapplies the momentum equation, leading to an incorrect speed after the ball bounces off. The correct approach involves using the conservation of momentum for both the ball and the person, taking into account the direction of the ball's velocity after the collision. Ultimately, the correct speed after the collision is determined to be 10.3 cm/s.
kevina
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The Question:

(1) You are standing on a sheet of ice that covers a parking lot; there is negligible friction between your feet and the ice. A friend throws you a 0.400 kg ball that is traveling horizontally at 10.2 m/s. Your mass is 73.4 kg.

(a) If you catch the ball, with what speed do you and the ball move afterward? Answer in cm/s.

(b) If the ball hits you and bounces off your chest, so that afterward it is moving horizontally at 8.70 m/s in the opposite direction, what is your speed after the collision? Answer in cm/s.

My Solution:
I was able to solve Part A. It's Part B I'm having trouble with. I used
(.400kg)(870cm/s)=(73.4kg)(v)
I solved for v and got 4.74cm/s, but the answer is wrong.
What am I doing wrong?
 
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kevina said:
I used
(.400kg)(870cm/s)=(73.4kg)(v)
That would be true if the total momentum were zero, but it's not.

Set the initial momentum of the ball equal to the final momentum of the ball and you.
(Careful with signs: the ball bounces off and goes in the negative direction.)
 
kevina said:
The Question:

(1) You are standing on a sheet of ice that covers a parking lot; there is negligible friction between your feet and the ice. A friend throws you a 0.400 kg ball that is traveling horizontally at 10.2 m/s. Your mass is 73.4 kg.

(a) If you catch the ball, with what speed do you and the ball move afterward? Answer in cm/s.

(b) If the ball hits you and bounces off your chest, so that afterward it is moving horizontally at 8.70 m/s in the opposite direction, what is your speed after the collision? Answer in cm/s.

My Solution:
I was able to solve Part A. It's Part B I'm having trouble with. I used
(.400kg)(870cm/s)=(73.4kg)(v)
I solved for v and got 4.74cm/s, but the answer is wrong.
What am I doing wrong?

Part B is an elastic collision.. and momentum is always conserved... the base eqaution is this... m1v1 (+/-) m2v2 = m1v3 (+/-) m2v4

v3 and v4 are the new velocities of the objects after the collision.

so it would be like this...

m1= ball
m2= you
v1=ball
v2= you
v3= ball after collision
v4= you after collision

m1v1 + 0 = -(m1v3) + (m2v4)

solving for v4 will equal

v4 = (m1v1) + (m1v3)
----------------
m2


=0.103m/s

=10.3 cm/s
 
Ahh.. I see.. I wasn't using the right equation. Thanks for the help Doc Al and hagar852. :)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

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