Six similar boxes (A–F) are initially sliding in the positive direction along a

Question

Six similar boxes (A–F) are initially sliding in the positive direction along a frictionless horizontal surface at a speed of 10 m/s. Then a net horizontal force, also in the positive direction, is applied to each box for a period of 10 seconds. The masses of the boxes and the net horizontal force for each case are given below.

Rank the boxes in order of increasing FINAL momentum. That is, put first the box with the smallest final momentum, and put last the box with the largest final momentum.

If B is smallest, then A, C, D, and finally E is largest, enter BACDE.
Note: if final momenta are equal, then enter those cases in the order listed.

A) F = 75 N. . . . . . M = 10 kg
B) F = 55 N. . . . . . M = 20 kg
C) F = 70 N. . . . . . M = 25 kg
D) F = 70 N. . . . . . M = 20 kg
E) F = 120 N. . . . . . M = 5 kg
F) F = 120 N. . . . . . M = 20 kg

Explanation / Answer

here,

initial velocity , u = 10 m/s

time period , t = 10 s

change in momentum = force * time

A)

let the final momentum be P

F = 75 N

M = 10 kg

change in momentum = force * time

P - M * u = F * t

P - 10 * 10 = 75 * 10

P = 850 kg . m/s

B)

let the final momentum be P

F = 55 N

M = 20 kg

change in momentum = force * time

P - M * u = F * t

P - 20 * 10 = 55 * 10

P = 750 kg . m/s

C)

let the final momentum be P

F = 70 N

M = 25 kg

change in momentum = force * time

P - M * u = F * t

P - 25 * 10 = 70 * 10

P = 950 kg . m/s

D)

let the final momentum be P

F = 70 N

M = 20 kg

change in momentum = force * time

P - M * u = F * t

P - 20 * 10 = 70 * 10

P = 900 kg . m/s

E)

let the final momentum be P

F = 120 N

M = 5 kg

change in momentum = force * time

P - M * u = F * t

P - 5 * 10 = 120 * 10

P = 1250 kg . m/s

F)

let the final momentum be P

F = 120 N

M = 20 kg

change in momentum = force * time

P - M * u = F * t

P - 20 * 10 = 120 * 10

P = 1400 kg . m/s

the boxes in order of increasing FINAL momentum are BADCEF

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