A rubber ball of mass 0.2 kg is released from an upstairs window of a building and strikes the ground at a velocity of 30 m/s. The ball bounces upward from the ground at a speed of 20 m/s. What is the impulse for the time the ball is in contact with the ground (in N s)?
Thank you!
2007-12-10 10:52:06 · 2 answers · asked by carrerasarah 2 in Science & Mathematics ➔ Physics
J = d(p) = F dt; where J is the impulse, dp is the change in momentum. Therefore, J = m(v1 - v0); where m = .2 kg, v0 = -30 mps, and v1 = 20 mps and velocity up is positive. Therefore, the impulse is J = .2(20 + 30) = 10 kg (m/sec) or 10 kg (m/sec^2) sec = 10 N-sec; where kg (m/sec^2) is a Newton.
2007-12-10 11:03:10 · answer #1 · answered by oldprof 7 · 0⤊ 0⤋
Impulse is the change in momemntum
change in momentum = mVf - mVi
suppose up is positive, then
mVi = 0.2(-30) why negative?
mVf = 0.2(20)
impulse = 0.2(20)-(0.2)(-30) = 1 Ns
2007-12-10 19:01:00 · answer #2 · answered by Anonymous · 0⤊ 1⤋