- #1
- 14
- 0
0.5Kg snowball moving at 20m/s strikes and sticks to a 70Kg man standing on the frictionless surface of a frozen pond. what is the final velocity?
No. To find the total momentum of the system, just add up the momentum of each part. What's the initial momentum of the snowball? Of the man?uhmm, the momentum before and after the collision was zero?
In the system of man and snowball total momentum is conserved. The total momentum before the collision is not zero, thus it cannot be zero after the collision. Could you write down the equations for momentum for the system before the collision? p=...?uhmm, the momentum before and after the collision was zero?
can i use this formula?
(0.5kg)*(20m/s)+(70kg)(0m/s)=(0.5kg+70kg)(Vafter)?
10kg.m/s=(70.5kg)(Vafter)
Vafter=10kg.m/s / 70.5kg
= 0.14m/s
tama ako means im correct... bye.. thanks :zzz:Tama ako? Anyway yes you found the correct method and answer to your problem.
oh! i see,, i'll keep that in mind... well thanks anyway,, i still have 10 more questions to answer tomorrow,, thank you,, nyters and byersIt's a matter of significant digits really.
The correct equation to use was [itex]m_{sb} v_{sb}+m_{man}v_{man}=(m_{sb}+m_{man}) v_{final}[/itex]. The interesting part is [itex](m_{sb}+m_{man}) =70.5kg[/itex].
The equation you used the first time was [itex]m_{sb} v_{sb}+m_{man}v_{man}=m_{man} v_{final}[/itex]. Here you are ignoring that when a snowball is stuck to a man his total mass is increased. Because the snowball is so light both equations gave the same answer with a 2-decimal accuracy. Try to recalculate it by saying that the mass of the snowball is 100kg, keep the rest of the numbers the same. You will see that your answer is very different.
More intuitively: If I throw a snowball at you, you won't move much because its mass is low. If I were to throw a wrecking ball at you, you would move quite a bit since its mass is high compared to yours, assuming both snowball and wrecking ball hit you at the same speed.