In tae-kwon-do, a hand is slammed down onto a target at a speed of 10 m/s and comes to a stop during the 7.0 ms collision. Assum
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Answer:
- After throwing the snow, velocity of the thrower is 2.33 m/s
- the velocity of the receiver is 0.026 m/s
Explanation:
Given the data in the question;
Using conservation of momentum,
Initial thrower has a momentum of mv; v
(69.5 kg + 0.0475 kg) × 2.35 m/s = 163.4366 kg.m/s
Now, When he throws it at 31.5 m/s, these constitutes a momentum of;
(0.0475 kg )(31.5 m/s) = 1.49625 kg.m/s
hence his momentum now is: 163.4366 - 1.49625 = 161.94035 kg.m/s
To get his velocity, we say;
161.94035 = mv
{ he lost weight of the snow ball so, m = 69.5 kg )
161.94035 = 69.5 × v
v = 161.94035 / 69.5
v = 2.33 m/s
Therefore, After throwing the snow, velocity of the thrower is 2.33 m/s
Next is the Receiver;
the receiver will gain momentum of 1.49625 kg.m/s
he has no momentum initially and after he catches the snow ball;
1.49625 kg.m/s = mv
1.49625 kg.m/s = ( 57.5 kg + 0.0475 kg ) × v
1.49625 kg.m/s = 57.5475 kg × v
v = ( 1.49625 kg.m/s ) / 57.5475 kg
v = 0.026 m/s
Therefore, the velocity of the receiver is 0.026 m/s