PLS EXPLAIN ME THE FIRST QUESTION (A)
- Thread starter Keloglan
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- #2
To determine the impulse(FΔt), you will want to know the velocity before (V) and the velocity after impact (V') or the change in momentum(Δp).
We know the ball is 0.5kg (m = 0.50 kg), the initial height is 7.2m (h = 7.2m) and it rebounds to 5m (h' = 5.0m), the time of contact with the floor (Δt = 0.1s) and we assume a gravitational acceleration of g = 10m/s²
FΔt = Δp = m(V+V') -- This is the mass multiplied by the change in velocity
So we just plug the values that we know into the formula.
V = √2gh
V' = √2gh
V = √2*10*7.2 = 12 m/s
V = √2*10*5 = 10 m/s
Now, for the impulse
FΔt = Δp = m(V+V')
FΔt = Δp = 0.50*(10+12) = 11
FΔt = 11 Ns
It has been quite a few years since I last did any of this, but I hope it made sense
We know the ball is 0.5kg (m = 0.50 kg), the initial height is 7.2m (h = 7.2m) and it rebounds to 5m (h' = 5.0m), the time of contact with the floor (Δt = 0.1s) and we assume a gravitational acceleration of g = 10m/s²
FΔt = Δp = m(V+V') -- This is the mass multiplied by the change in velocity
So we just plug the values that we know into the formula.
V = √2gh
V' = √2gh
V = √2*10*7.2 = 12 m/s
V = √2*10*5 = 10 m/s
Now, for the impulse
FΔt = Δp = m(V+V')
FΔt = Δp = 0.50*(10+12) = 11
FΔt = 11 Ns
It has been quite a few years since I last did any of this, but I hope it made sense
