here at the moment when bullet strikes the block it will have no external force on it
so here we can use momentum conservation
so we will have
so they move together with speed 1.98 m/s
2.10-kg box is moving to the right with speed 8.50 m>s on a horizontal, frictionless surface. At t = 0 a horizontal force is
applied to the box. The force is directed to the left and has magnitude F1t2 = 16.00 N>s22t2. (a) What distance does the box move from its position at t = 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at t = 3.00 s?
2 answers:
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Answer:
k = 1/18
Explanation:
Data:
a = -k
to = 0s Vo = 12m/s
t = 6s the particle chage it's moviment, so v = 0 m/s
We know that acceleration is the derivative of velocity related to time:
rearranging...
Then, we must integrate both sides:
V = 0 because the exercise says that the car change it's direction:
k = 1/6
In order to find X - Xo we must integer v*dT = dX
so...
integrating...
X - Xo = 54m