Question

A 200 kg golf cart strikes a 1500 kg car with 1000N of force, bringing its velocity from 5 m/s to 0 m/s in 1 second. How much force did the car apply to the golf cart?
Question 6 options:

100 N


1000 N


1500 N


Not enough information has been given to solve the problem

Answer 1

Answer:

1000N

Explanation:

From Newton's third law of motion, for every action there is an equal and opposite reaction. In other words, when two bodies are in collision, the forces acting between these bodies are equal in magnitude and opposite in direction.

In the case presented in the question, the bodies are the golf cart and the car. During collision, the golf car strikes the car with a 1000N of force (that is the action). Consequential of this, though in the opposite direction, the car will apply an equal force of 1000N (that is the reaction) on the golf car.

Therefore the car applied a force of 1000N on the golf cart.

Answer 2

Answer:

The car applied a force of 1000 N to the golf cart

Explanation:

Based on the laws of conservation of linear momentum, we have that linear momentum is always conserved during a collision.

By Newtons Second law, (impulse law),

$Ft = [m_{1}(v-u)+ m_{2}(v-u)]$

Let m 1 be the mass of the golf cart, and v 1 its initial velocity. They are = 200 kg and 5 m/s respectively.

Let m 2 be the mass of the car, and  V 2 it initial velocity. They are = 1500 kg and 0 m/s respectively.

The time involved is 1 second

Applying the values of the variables to the formula, we have.

$F \times 1 = [200 \times (5-0) - 1500 \times (0-0)]$

From this we have that the force the car applied to the golf cart is = 1000 N