Respuesta :
Answer:
The final velocity of the bumper car is 1.2 m/s in the direction opposite to the initial motion direction.
Explanation:
Given:
Mass of the car and driver (M) = 200 kg
Force exerted by the side rail (F) = -4000 N (Negative sign implies the force is in opposite direction)
Time for which force acts (t) = 0.200 s
Initial velocity of the car (u) = 2.80 m/s
Final velocity of the car (v) = ?
Now, in order to determine the final velocity, we first find impulse acting on the bumper by the side rail.
Impulse is given as the product of force and time interval. So,
Impulse, [tex]J=Ft=-4000\times0.200=-800\ Ns[/tex]
Now, we know that, impulse is nothing but change in momentum of the body.
So, change in momentum of the bumper is given as:
[tex]\Delta P = P_f-P_i[/tex]
Where,
[tex]P_i\to initial\ momentum=Mu\\P_f\to final\ momentum=Mv[/tex]
Therefore, [tex]\Delta P=Mv-Mu=200v-(200\times 2.80)=(200v-560)\ Ns[/tex]
Now, Change in momentum = impulse
⇒ [tex]200v-560=-800[/tex]
⇒ [tex]200v=-800+560[/tex]
⇒ [tex]v=\frac{-240}{200}=-1.2\ m/s[/tex]
Negative sign implies the direction of the bumper after collision is opposite to that of the initial direction before collision.
Therefore, the final velocity of the bumper car is 1.2 m/s in the direction opposite to the initial motion direction.
