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Momentum Word Problems
Ex 1) A 5.0 kg mass has its velocity change from 8.0 m/s east to 2.0 m/s east. Find the objects change in momentum.
Ex 1) A 5.0 kg mass has its velocity change from 8.0 m/s east to 2.0 m/s east. Find the objects change in momentum. m = 5.0 kg
Δp = mΔV
= (5.0 kg)(2.0 m/s - 8.0 m/s)
= -30. kgm/s East
or +30. kg m/s West
Ex 2) A 5.0 kg mass moving with a velocity of 8.0 m/s east has an impulse applied to it which causes its velocity to change to 20. m/s East.
Find Impulse:
Ex 2) A 5.0 kg mass moving with a vector of 8.0 m/s east has an impulse applied to it which causes its velocity to change to 20. m/s East. Find Impulse: m = 5.0 kg
J = mΔv = (5.0 kg)(12. m/s East) = 60. kg m/s east = 60. Ns East
Find the force if the impulse was applied for 3.0 sec. m = 5.0 kg
J = Ft = 60. Ns
F(3.0 sec) = 60. Ns
F = 20 N East
Ex 3) How long would it take for a net upward force of 100. N, to increase the speed of a 50. kg object from 100. m/s to 150. m/s.
Ex 3) How long would it take for a net upward force of 100. N, to increase the speed of a 50. kg object from 100. m/s to 150. m/s.
F = 100. N m = 50. kg Vi = 100. m/s East Vf = 150. m/s East t = ?
Ft = mΔv
(100. N)t = 50. kg(50. m/s)
t = 25. seconds
Ex 4) A 1.0 kg ball traveling @ 4.0 m/s strikes a wall and bounces straight back @ 2.0 m/s.
Find Δp
Ex 4) A 1.0 kg ball traveling @ 4.0 m/s strikes a wall and bounces straight back @ 2.0 m/s. Find Δp
m = 1.0 kg Vi = 4.0 m/s East Vf = ????? Vf = -2.0 m/s (opposite direction) Δp = ?
(a) Δp = mΔv = (1.0 kg)(-2.0 m/s - 4.0 m/s) = - 6.0 kgm/s
(b)
What is impulse applied to the ball?
J = Δp = - 6.0 kgm/s
(c) What is impulse applied to the wall?
J = + 6.0 kgm/s
3rd Law, Action Reaction
Law Of Conservation Of Momentum
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