Calculating Time of Impact for a Ball Thrown Against a Wall

  • Thread starter Thread starter SamAnthony
  • Start date Start date
  • Tags Tags
    Impact Time
AI Thread Summary
To calculate the time of impact for a ball thrown against a wall, the mass must be converted from grams to kilograms, as this is essential for accurate calculations. The ball, with a mass of 0.05 kg and a velocity of 12 m/s, exerts a force of 3 N on the wall. The change in momentum is calculated using the formula Δp = F * dt, which leads to the determination of time of impact. There is some confusion regarding whether the collision is elastic, but it is noted that this detail may not be necessary for the calculation. Overall, the discussion emphasizes the importance of proper unit conversion and understanding the relationship between force, momentum, and time.
SamAnthony
Messages
4
Reaction score
0

Homework Statement


A ball of mass 50g is thrown horizontally against a wall such that it strikes it
at a velocity of 12ms-1. If it exerts a force of 3N on the wall, determine the
time of impact.

Homework Equations


The Attempt at a Solution



I drew the diagram and made the direction and wrote the values neatly on my paper as it's sort of a must to do this and it makes me understands the work better. Does the 50g needs to be changed and converted into kg?
 
Physics news on Phys.org
is the collision elastic??
 
as we know
dp=F.dt
and hence you could find the answer
 
SamAnthony said:
Does the 50g needs to be changed and converted into kg?


yes it is needed to...!
 
I got totally confused! that's the full question. So is it 0.2s?
 
i think the question is some how incomplete... please look and write all the information given in the question...!
 
That's the full question! maybe there's a trick in it? not sure, otherwise there's another value which could be taken as 0?
 
then it will be . change in momentum = Δp = p(initial)-p(final)
assume the ball is movin along +ve x direction
then
Δp = p-(-p)= 2p
an hence pu it in the above equation
 
  • Like
Likes 1 person
click thanks on it...:)
 
  • #10
Sure! :) I'm new to this, and I'm figuring this forum out! and thanks for your help too
 
  • #11
he he he ok...!
 
  • #12
Kishlay said:
is the collision elastic??
you don't need to know. Change in momentum = ∫F.dt.
 
Back
Top