A ball whose mass is 100g is dropped from a height of 2m from the floor. It rebounds vartically upward after colliding with the floor to the height 1.5m . find (a) the momentum of a the before and after colliding with the floor (b) the average force exerted by the floor on the Ball . Assume that collision lasts for -10×8
Answers
Answered by
30
V2−U2 = 2gS
Here given that, U = 0 m/s, S = 1.5 m. So,
V2−02 = 2×10×2 ⇒V=40−−√ m/s.
So, momentum of the ball is,
Pi=mV = 0.1×40−−√ = 0.1×6.325 = 0.633⇒ Pi=0.633 kg.m/s
After collision, it is given that the ball rises to a height S' = 1.5 m.
So, velocity with which the body rises is,
V2−U2=−2gS' ⇒ 02 − V'2 = −2×10×1.5 ⇒V' = 30−−√ m/s.
So, the momentum of the ball, when the ball is bouncing back.
Pf=mV' = 0.1×30−−√⇒Pf=0.547 kg m/s.
Therefore, the average force exerted by the floor on the ball is,
Favg=∆PDt = 0.633−0.54710−8=0.086×108 ⇒Favg=8.6×10^6N
Here given that, U = 0 m/s, S = 1.5 m. So,
V2−02 = 2×10×2 ⇒V=40−−√ m/s.
So, momentum of the ball is,
Pi=mV = 0.1×40−−√ = 0.1×6.325 = 0.633⇒ Pi=0.633 kg.m/s
After collision, it is given that the ball rises to a height S' = 1.5 m.
So, velocity with which the body rises is,
V2−U2=−2gS' ⇒ 02 − V'2 = −2×10×1.5 ⇒V' = 30−−√ m/s.
So, the momentum of the ball, when the ball is bouncing back.
Pf=mV' = 0.1×30−−√⇒Pf=0.547 kg m/s.
Therefore, the average force exerted by the floor on the ball is,
Favg=∆PDt = 0.633−0.54710−8=0.086×108 ⇒Favg=8.6×10^6N
Answered by
10
The change in momentum is a definition correct.
But direction of momentum is different therefore its getting added.
So delta P=0.626+0.542
And force will be 1.168* 10^8
Similar questions