Learning Goal: To apply the principle of linear impulse and momentum to a mass t

Question

Learning Goal:

To apply the principle of linear impulse and momentum to a mass to determine the final speed of the mass.

A 10-kg, smooth block moves to the right with a velocity of v0 m/s when a force F is applied at time t0=0 s.

(Figure 1)

Part A - Constant forces

At time t0=0 s, a force is applied to the block. If the block is moving with an initial velocityv0=2.25 m/s and the force varies as shown in the graph, (Figure 2) , wheret1=2 s, t2=4 s,and t3=6 s, what is the speed of the block at time t3? v(6 s)= ?

Express your answer to three significant figures.

Part B - The speed of the block at t 3

At time t0=0 s, a force is applied to the block. If the block is moving with an initial velocityv0=4.5 m/s and the force varies as shown in the graph, (Figure 3) , wheret1=1.25 s,t2=2.5 s, and t3=3.75 s, what is the speed of the block at time t3? v(3.75 s)= ?

Express your answer to three significant figures.

Part C - The time it take to stop the motion of the block

At time t0=0 s, a force is applied to the block. The force varies as shown in the graph, (Figure 4) ,where t1=1.25 s, t2=2.5 s, and t3=3.75 s. If the block is moving with an initial velocityv0=3.5 m/s and the force remains constant at ?20 N for all times greater than t3, at what timetf does the block stop moving to the right?

Express your answer to three significant figures.

fugures are in order 1,2,3,4

Explanation / Answer

A. Area under the F versus t curve will give change in momentum.

Area = (20 x t1) + (10 x (t2 - t1)) + (-10 x (t3 -t2))

    = (20 x 2) + (10 x 2) + (-10 x 2) = 40 kg m/s

Change in momentum = mass ( vf - vi )

40 = 10 (vf - 2.25)

vf = 6.25 m/s

B.

Area = (20 x t2 / 2 ) + (-20 x (t3 -t2) / 2)

     = (20 x 2.5 / 2) + (-20 x (3.75 - 2.5) / 2) = 12.5 kg m/s

change in momentum = 12.5 kg m /s

12.5 = 10 (vf - 4.5)

vf = 5.75 m/s

C.

Initial momentum = mv0 = 10 x 3.5 = 35 kg m/s

when area is equal to - 35 then block will stop.

Area = (20 x 1.875 / 2 ) + (-20 x 0.625 / 2) + (-20 x (t - 2.5))

-35 = 18.75 - 6.25 - 20t + 50

t = 4.875 sec

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