Respuesta :
a) The average force exerted on car B by car A is 3 N
b) The average force exerted on car A by car B is 3 N (in the opposite direction)
c) The velocity of toy car A after the collision is 0.7 m/s
Explanation:
a)
We can calculate the average force exerted on car B during the collision by applying Newtons' second law, which states that the force exerted on an object is equal to the product between its mass and its acceleration:
[tex]F=ma[/tex]
where, in this case,
F is the average force exerted on car B
m is the mass of car B
a is the acceleration of car B
The data that we have in this problem are
m = 1 kg is the mass of car B
[tex]a=3 m/s^2[/tex] is the acceleration of car B during the collision
Substituting,
[tex]F=(1)(3)=3 N[/tex]
b)
We can answer this second question by applying Newton's third law of motion, which states that:
"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"
In this situation, we can call:
- Action the force exerted by car A on car B
- Reaction the force exerted by car B on car A
According to Newton's third law, action and reaction are equal in magnitude and opposite in direction: therefore, since the action is 3 N (calculated in part a), the reaction force must also be 3 N. So, the force exerted by car B on car A is also 3 N, in the opposite direction.
3)
We can solve this part by applying the impulse theorem and the law of conservation of momentum. In fact:
The change in momentum of car B during the collision is equal to the impulse exerted on it, therefore
[tex]\Delta p_B = F \Delta t[/tex]
where
F = 3 N is the average force exerted on B
[tex]\Delta t = 0.5 s[/tex] is the duration of the collision
Substituting,
[tex]\Delta p_B = (3)(0.5)=1.5 kg m/s[/tex]
The initial momentum of car A before the collision is
[tex]p_A = m_A u_A = (3 kg)(1.2 m/s)=3.6 kg m/s[/tex]
where [tex]m_A[/tex] is the mass of car A and [tex]u_A[/tex] its initial velocity.
According to the law of conservation of momentum, the impulse gained by car B is equal to the impulse lost by car A, therefore:
[tex]p_A - p_A' = \Delta p_B[/tex]
where [tex]p_A'[/tex] is the final momentum of car A, after the collision. Solving,
[tex]p_A'=p_A - \Delta p_B = 3.6-1.5 =2.1 kg m/s[/tex]
And the final momentum of A can be written as
[tex]p_A' = m_A v_A[/tex]
where [tex]v_A[/tex] is its velocity after the collision. Solving for [tex]v_A[/tex],
[tex]v_A = \frac{p_A'}{m_A}=\frac{2.1}{3}=0.7 m/s[/tex]
Learn more about Newton laws of motion and momentum:
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