A Snowball in Snowfall

Can someone please point out the flaw in my working out.
https://isaacphysics.org/questions/snowball?stage=a_level
I thought the time was 15/g as after this the snowball would go faster than the snowfall.

Working:

Original post by mosaurlodon

Working:

The final time would be when the ball is falling at the same speed as the snow, not when the ball is stationary. So it would be a simple function of vs at well.

Reply 3

mosaurlodon

Thank you very much, i've got the answer.
Quick q tho: shouldnt the factor of snow accumulation change from vs+u/vs to u-vs/vs since its starting to move down from the top?

Reply 4

mqb2766

Original post by mosaurlodon

Thank you very much, i've got the answer.
Quick q tho: shouldnt the factor of snow accumulation change from vs+u/vs to u-vs/vs since its starting to move down from the top?

No, its really the relative velocity (difference in velocities) thats important. Here the difference is skated over when they add the two speeds as theyre heading in different directions (at least initially). The relative velocity is vs+v (so 16 initially) until v = -vs (so both descending at 1m/s). If theyre falling at the same rate, the snow will never accumulate on the ball. Personally, the formula would make more sense if it was the difference and both terms were velocities with the same +ive direction. In the diagram (hint 5), the peak height of the ball would correspond to the top cylinder being 0 volume. As it begins to descend, the top cylinder would have negative volume until theyre falling at the same rate and top cylinder would have the same (negative) volume as the bottom one and they cancel.

It could be argued that on the way down (so speed > 1m/s) the ball would accumulate snow on the bottom as the ball is travelling faster than the snow is falling, but thats (obviously) not in the scope of the question as the hint gives.

(edited 1 month ago)

Reply 5

mosaurlodon

Thank you for clarifying.
"It could be argued that on the way down (so speed > 1m/s) the ball would accumulate snow on the bottom as the ball is travelling faster than the snow is falling" - wouldnt this snow already be picked up by the ball on its way up.

Reply 6

mqb2766

Original post by mosaurlodon

probably.

Reply 7

Hihobyu

Original post by mqb2766

The final time would be when the ball is falling at the same speed as the snow, not when the ball is stationary. So it would be a simple function of vs at well.

Would the final time t=(v-15)/g because of v =u + at??

Reply 8

mosaurlodon

it should be v+15/g
the velocity directions are a bit weird with this one but if you imagine the time the snowball takes to go up until stationary.
which is 15/g (from my working out image)

then you work out time the snowball takes to go from stationary to v_s
which is v_s/g (since time is always positive)

then you add them - hope this helps