In: Physics
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 = 30 N. . . . . . M = 15 kg
B) F = 80 N. . . . . . M = 10 kg
C) F = 70 N. . . . . . M = 15 kg
D) F = 75 N. . . . . . M = 15 kg
E) F = 95 N. . . . . . M = 25 kg
F) F = 110 N. . . . . . M = 10 kg
Six similar boxes (A–F) are initially sliding in the positive direction along a frictionless horizontal surface.
Given that :
velocity of the box, v = 10 m/s
time taken by each box, t = 10 sec
According to newton's law, Force = rate of change in momentum
F = (p2 - p1) / t { eq.1 }
rearranging an above eq.
Ft = p2 - p1
p2 = Ft + p1
where, p1 = initial momentum = m1 v
then, p2 = Ft + m1 v { eq.2 }
Fina lmomentum for each box which is given as ::
(A) F = 30 N. . . . . . M = 15 kg
p2 = (30 N) (10 s) + (15 kg) (10 m/s)
p2 = 450 kg.m/s
(B) F = 80 N. . . . . . M = 10 kg
p2 = (80 N) (10 s) + (10 kg) (10 m/s)
p2 = 900 kg.m/s
(C) F = 70 N. . . . . . M = 15 kg
p2 = (70 N) (10 s) + (15 kg) (10 m/s)
p2 = 850 kg.m/s
(D) F = 75 N. . . . . . M = 15 kg
p2 = (75 N) (10 s) + (15 kg) (10 m/s)
p2 = 900 kg.m/s
(E) F = 95 N. . . . . . M = 25 kg
p2 = (95 N) (10 s) + (25 kg) (10 m/s)
p2 = 1200 kg.m/s
(F) F = 110 N. . . . . . M = 10 kg
p2 = (110 N) (10 s) + (10 kg) (10 m/s)
p2 = 1200 kg.m/s
Rank the boxes in order of increasing FINAL momentum. Put first the box with the smallest final momentum & Put last the box with the largest final momentum.
A C B D E F