Question

6. A 500-gram puck and a 250-gram puck collide on a frictionless air table. During the collision,

Answer 1

a) The force acting on the 250-gram puck is 1.0 N

b) The force acting on the 500-gram puck is 1.0 N (in the opposite direction)

c) The acceleration of the 500-gram puck is $2.0 m/s^2$

Explanation:

a)

The force acting on the 250 gram puck can be found by using Newton's second law, which states that the net force on an object is equal to the product between its mass and its acceleration:

$F=ma$

where:

F is the net force

m is the mass of the object

a is the acceleration

For the 250-gram puck,

m = 250 g = 0.250 kg

$a=4.0 m/s^2$

So the net force on this puck is

$F=(0.250)(4.0)=1.0 N$

b)

We can solve this part by using Newton's third law, which states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

In this situation, we can identify the 250 g puck as object A, and the 500 g puck as object B.

This means that if the force exerted by the 500-g puck on the 250-g puck is 1.0 N (calculated in part a), therefore the force exerted by the 250-g puck on the 500-g puck must also be 1.0 N (but in the opposite direction).

c)

We can solve also this part by using Newton's second law. In fact:

$F=ma$

where:

F is the net force

m is the mass of the object

a is the acceleration

For the 500-gram puck,

m = 500 g = 0.500 kg

$F=1.0 N$

So we can re-arrange the equation to find the acceleration:

$a=\frac{F}{m}=\frac{1.0}{0.500}=2.0 m/s^2$

Learn more about Newton laws of motion:

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