What is the velocity of a 900-kg car initially moving at 30.0 m/s, just after it hits a 150-kg deer initially running at 18.0 m/s in the same direction? Assume the deer remains on the car.
28.29m/s
In this situation, linear momentum is conserved. And since the deer remains on the car after collision, the linear momentum is given as;
([tex]m_{C}[/tex] x [tex]u_{C}[/tex]) + ([tex]m_{D}[/tex] x [tex]u_{D}[/tex]) = ([tex]m_{C}[/tex] + [tex]m_{D}[/tex]) v -----------------(i)
Where;
[tex]m_{C}[/tex] = mass of car
[tex]u_{C}[/tex] = initial velocity of car before collision
[tex]m_{D}[/tex] = mass of deer
[tex]u_{D}[/tex] = initial velocity of the deer before collision
v = common velocity with which the car and the deer move after collision
From the question;
[tex]m_{C}[/tex] = 900kg
[tex]u_{C}[/tex] = +30.0m/s (direction of the motion of the car taken positive)
[tex]m_{D}[/tex] = 150kg
[tex]u_{D}[/tex] = +18.0m/s (relative to the direction of the car, the velocity of the deer is also positive )
Substitute these values into equation (i) as follows;
(900 x 30.0) + (150 x 18.0) = (900 + 150)v
27000 + 2700 = 1050v
29700 = 1050v
v = [tex]\frac{29700}{1050}[/tex]
v = 28.29m/s
Therefore, the velocity of the car after hitting the deer is 28.29m/s. This is also the velocity of the deer after being hit by the car.
