A teacher is demonstrating acceleration to his class by throwing two balls. He throws both balls using the same amount of force in the same direction. The first ball he throws has a mass of 0.1 kg and the second ball has a mass of 0.5 kg. Explain why.

Responses

The balls will have equal acceleration because they are being thrown with the same force.
The balls will have equal acceleration because they are being thrown with the same force.

The first ball has a smaller mass, so it will have less acceleration than the second ball.
The first ball has a smaller mass, so it will have less acceleration than the second ball.

The first ball has a smaller mass, so it will have greater acceleration than the second ball.
The first ball has a smaller mass, so it will have greater acceleration than the second ball.

The balls will have equal acceleration because they are being thrown in the same direction.
The balls will have equal acceleration because they are being thrown in the same direction.

The correct explanation is:

The first ball has a smaller mass, so it will have greater acceleration than the second ball.

This is based on Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The formula for this relationship is F = ma, where F is the force applied, m is the mass of the object, and a is the acceleration.

Given that the same force is applied to both balls, the ball with the smaller mass (0.1 kg) will undergo a greater acceleration because the force is distributed over less mass. In contrast, the ball with the larger mass (0.5 kg) will have a smaller acceleration since the same force is now spread over a greater mass.

Mathematically, if we set the force F to be constant, we can compare accelerations (a) of ball 1 (m1 = 0.1 kg) and ball 2 (m2 = 0.5 kg):

a1 = F / m1
a2 = F / m2

Since m1 < m2, it follows that a1 > a2, meaning the smaller mass leads to a larger acceleration if the force is constant.