The forces shown in the force vs. time diagram in the figure below act on a 1.6-
ID: 1260124 • Letter: T
Question
The forces shown in the force vs. time diagram in the figure below act on a 1.6-kg particle.
(a) Find the impulse for the interval from
t = 0
to
t = 3.0 s.
(b) Find the impulse for the interval from
t = 0
to
t = 5.0 s.
(c) If the forces act on a 1.6-kg particle that is initially at rest, find the particle's speed at
t = 3.0 s.
(d) If the forces act on a 1.6-kg particle that is initially at rest, find the particle's speed at
t = 5.0 s.
The forces shown in the force vs. time diagram in the figure below act on a 1.6-kg particle. (a) Find the impulse for the interval from t = 0 to t = 3.0 s. (b) Find the impulse for the interval from t = 0 to t = 5.0 s. (c) If the forces act on a 1.6-kg particle that is initially at rest, find the particle's speed at t = 3.0 s. (d) If the forces act on a 1.6-kg particle that is initially at rest, find the particle's speed at t = 5.0 s.Explanation / Answer
The impulse is
I=integral (F)dt
Which is simply the area under the F vs. t diagram
(a).
I= area = height*width = 9N *0.3s = 2.7 Ns
(b)
The impulse for this one is the area for question (a) minus the area from t=0.3 to t=0.5 s
I = area(t=0~0.3) - area(t=0.3~0.5)
=2.7 - height * width
=2.7 - 4.5 N * 0.2 s
=2.7 - 0.9 = 1.8 Ns
(c)
One useful thing with Impulse is that it is equal to the change in momentum
I=delta p = (m*v2 - m*v1)
the particle is initially at rest, so v1=0
at t=0.3 s
delta p=(m *v2 - 0)
=1.7 kg *v2 = 2.7 Ns
v2=2.7 Ns/ 1.7 kg = 1.59 m/s
at t=0.5 s
delta p= m*v3 - m*v2
= 1.7* v3 - 1.7 * 1.59
=1.7* v3 - 2.7 = -0.9 Ns (-0.9 Ns is the impulse during this period)
v3=(-0.9+2.7)/1.7
= 1.06 m/s
Hope this help.
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