Two cars are parked on a street. A third car hits the back of the second car and causes all of the cars to move. (Ignore the distance between the cars). Calculate:
(a)The acceleration of the whole system;
(b)the net force on each car;
(c)the contact force that each car exerts on the neighbor;
(d)if all the masses are equal and 1200kg., and F=960 Newtons, give numerical answers to the previous parts.
You can not find the forces or acceleration without more information. The only thing you can use is the First Law, conservation of momentum, which will give you final speed if they stick together. (1/3 the original speed)
To answer questions about force and acceleration, you must know the time duration of the crash. ( change in momentum = integral of F dt)
Where did you get this question? Didd you make it up yourself or is there more information that you have not typed. Your other question about the mountain climber is also unanswerable as stated.
To calculate the answers to the questions, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = m*a).
Given:
Mass of each car (m) = 1200 kg
Force applied on the second car (F) = 960 N
(a) The acceleration of the whole system:
To calculate the acceleration of the whole system, we need to find the total force acting on the system and divide it by the total mass.
Let's assume all three cars move together as a single system. Since the third car hits the second car, the force applied on the second car will be the same magnitude as the force exerted by the third car.
The total force acting on the system can be found by adding up the forces acting on each individual car:
F_total = F_second_car + F_third_car
Since the third car hits the second car, the force applied on the second car will be directed towards the first car, and the force on the third car will be directed towards the second car.
F_total = F_second_car + (-F_third_car)
Given:
F_second_car = F = 960 N
F_third_car = -F = -960 N
F_total = 960 N + (-960 N) = 0 N
The total force acting on the system is zero. As a result, the acceleration of the whole system will be zero.
(b) The net force on each car:
Since the acceleration of the whole system is zero, the net force on each car will also be zero. This is because for a system to have zero acceleration, the sum of all forces acting on it must be zero.
(c) The contact force that each car exerts on the neighbor:
Since the net force on each car is zero, the contact force that each car exerts on the neighbor will also be zero. In this scenario, the forces between the cars cancel out, resulting in no net force.
(d) Numerical answers:
(a) The acceleration of the whole system = 0 m/s^2
(b) The net force on each car = 0 N
(c) The contact force that each car exerts on the neighbor = 0 N