Answer:
2.5 m/s east
Explanation:
Let east be the positive direction for velocity.
The change in momentum of the 0.75 kg model car is ...
m1·v2 -m1·v1 = (0.75 kg)(11 m/s) -(0.75 kg)(-9 m/s)
= (0.75 kg)(20 m/s) = 15 kg·m/s
The change in momentum of the 2.0 kg model car is the opposite of this, so the total change in momentum is zero.
m2·v2 -m2·v1 = (2 kg)(v2 m/s) -(2 kg)(10 m/s) = 2(v2 -10) kg·m/s
The required relation is ...
15 kg·m/s = -2(v2 -10) kg·m/s
-7.5 = v2 -10 . . . . divide by -2
2.5 = v2 . . . . . . . add 10
The velocity of the model truck after the collision is 2.5 m/s east.
Answer:
25 m/s
Explanation:
First of all, we can find the acceleration the object by using Newton's second law of motion:

where
F = 20.0 N is the net force applied on the object
m = 4.0 kg is the mass of the object
a is its acceleration
Solving for a, we find

Now we know that the motion of the object is a uniformly accelerated motion, so we can find its final velocity by using the following suvat equation:

where
v is the final velocity
u = 0 is the initial velocity
is the acceleration
t = 5 s is the time
By substituting,

Less than or equal to the magnitude of the vector