A 1200 kg car moving at 35kph runs into and sticks to a parked 1000kg car. the collision takes 0.15s and after the crash the car's wheels lock and they skid to a stop on the mu= 0.40 road. the final speed of both cars is 5.3m/s
a) find the distance tranvelled before skidding to a stop (3.6m)
b) the impact force on both cars (35000N)
c) the energy absorbed (26kJ)
Are the numbers in parentheses known correct answers or you own calculated results?
Help will be provided if work is shown
To solve this problem, we will use the principles of collisions, Newton's laws of motion, and the concept of work and energy.
a) To find the distance traveled before skidding to a stop, we can use the equation of motion:
v² = u² + 2as
where
v = final velocity (5.3 m/s)
u = initial velocity (35 kph = 35 * 1000/3600 m/s)
a = acceleration (given by the friction force, which is equal to μmg, where μ = 0.40 and m is the total mass of the two cars)
s = distance
First, let's convert the initial velocity from kph to m/s:
u = 35 * 1000/3600 m/s = 9.7 m/s
Now, we can substitute the given values into the equation:
5.3² = 9.7² + 2 * μ * m * s
Simplifying the equation and solving for s:
28.09 - 94.09 = 0.8m * s
-66 = 0.8s
s ≈ -82.5 m
Since distance cannot be negative, we take the absolute value:
s ≈ 82.5 m
Therefore, the distance traveled before skidding to a stop is approximately 82.5 meters.
b) To find the impact force on both cars, we can use the equation for impulse:
J = F * t
where
J = change in momentum (since the cars stick together, the total initial momentum is equal to the total final momentum)
F = impact force (what we want to find)
t = time of collision (0.15 s)
The initial momentum of the system is given by the sum of the individual momenta of the two cars:
initial momentum = m_1 * v_1 + m_2 * v_2
= 1200 kg * 9.7 m/s + 1000 kg * 0 m/s
= 11640 kg*m/s
The final momentum of the system is given by the total mass of the two cars (since they stick together) multiplied by the final velocity (which is 5.3 m/s):
final momentum = (1200 kg + 1000 kg) * 5.3 m/s
= 10600 kg*m/s
Since the initial and final momentum are equal, we have:
F * t = final momentum - initial momentum
F * 0.15 s = 10600 kg*m/s - 11640 kg*m/s
F * 0.15 s ≈ -1040 kg*m/s
Since the impact force cannot be negative, we take the absolute value:
F ≈ 1040 kg*m/s / 0.15 s
F ≈ 6933.33 N
Therefore, the impact force on both cars is approximately 6933.33 Newtons.
c) To find the energy absorbed, we can use the work-energy principle:
Work = ΔKE
where ΔKE is the change in kinetic energy. The initial kinetic energy of the system is given by the sum of the individual kinetic energies of the two cars:
initial KE = (1/2) * m_1 * v_1² + (1/2) * m_2 * v_2²
= (1/2) * 1200 kg * (9.7 m/s)² + (1/2) * 1000 kg * 0 m/s
= 55980 J
The final kinetic energy of the system is given by the kinetic energy of the two cars combined when they have a final velocity of 5.3 m/s:
final KE = (1/2) * (1200 kg + 1000 kg) * (5.3 m/s)²
= 14145 J
Therefore, the energy absorbed during the collision is given by the change in kinetic energy:
Energy absorbed = final KE - initial KE
= 14145 J - 55980 J
= -41835 J
Since energy cannot be negative, we take the absolute value:
Energy absorbed ≈ 41835 J
Therefore, the energy absorbed during the collision is approximately 41835 Joules.