two cars weighing 1500 kg are made to collide with a wall.the initial and final velocities are 15 m/s and 2.6 m/s.if the collision last for 0.15 seconds,then find impulsive force exerted on the car
To find the impulsive force exerted on the car, we can use the impulse-momentum principle. The impulse experienced by an object is equal to the change in momentum it undergoes during a collision. Mathematically, the impulse can be calculated as the product of the force applied and the time duration of the collision.
The impulse (J) is given by:
J = Δp = m * (vf - vi)
Where:
J = Impulse
Δp = Change in momentum
m = Mass of the car
vi = Initial velocity
vf = Final velocity
First, we need to find the change in momentum (Δp) of the car. The change in momentum is given by the difference between the final and initial momentum:
Δp = m * (vf - vi)
We know the initial velocity (vi) is 15 m/s, the final velocity (vf) is 2.6 m/s, and the mass (m) is 1500 kg. Plug these values into the equation:
Δp = 1500 kg * (2.6 m/s - 15 m/s)
Next, calculate the change in momentum:
Δp = 1500 kg * (-12.4 m/s)
Δp = -18600 kg·m/s
The negative sign indicates that the momentum is in the opposite direction of the initial velocity.
Now, we can calculate the impulse (J) by multiplying the change in momentum (Δp) by the time duration of the collision (t):
J = Δp * t
Given that the collision lasts for 0.15 seconds, we have:
J = -18600 kg·m/s * 0.15 s
J = -2790 N·s
Therefore, the impulsive force exerted on the car is -2790 N·s, where the negative sign represents that the force is in the opposite direction to the motion of the car.