Momentum Word Problems

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Momentum Word Problems

J = Ft = mΔv = Δp

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
V_i = 8.0 m/s East
V_f = 2.0 m/s East
Δp = ?

J = Ft = mΔv =Δp

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
V_i = 8.0 m/s East
V_f= 20. m/s East
J = ?

J = Ft = mΔv =Δp

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.

F = ?

t = 3 seconds

m = 5.0 kg

V_i = 8.0 m/s East

V_f = 20. m/s East

J = 60. kg m/s east

J = Ft = mΔv =Δp

J = Ft = 60. Ns East

F(3.0 sec) = 60. Ns East

F = 20. N East

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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

V_i = 100. m/s

V_f = 150. m/s

t = ?

J = Ft = mΔv =Δp

FΔt = mΔv

(100. N)t = 50. kg(50. m/s)

t = 25. secs

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

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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

V_i = 4.0 m/s

V_f = ?

V_f = -2.0 m/s

(opposite direction)

Δp = ?

J = Ft = mΔv =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

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AP Physics

p. 188) 17

Dp = mDv

Momentum only changes in the x direction

toward wall +

V_i = Vsinq

V_f = -Vsinq

Dp = mDv

Dp = m(-2Vsinq)

= 2.1 kgm/s left

[Law Of Conservation

Of Momentum]