Question

F = m x a. A ball of mass 0.35kg rolls on a smooth surface hits another object at rest and of mass 7kg. The second object gets an acceleration of 1.5m/s2. What acceleration does the ball get? Use newton’s third law.

Please show the working and thankyouu!!

Answer 1

The acceleration the ball gets = 30 m/s²

Step-by-step explanation:

The given parameters are;

The mass of the first ball that hits an object, m₁ = 0.35 kg

The mass of the object hit by the ball, m₂ = 7 kg

The acceleration given to the second object, a₂ = 1.5 m/s²

Let the a₁ represent the acceleration the ball gets

Newton's third law states that action and reaction are equal and opposite, therefore, the force exerted on a body A by a body B is equal to the force exerted on the body B by the body A;

Therefore, we have;

The force exerted on the object, F₂ = m₂ × a₂ = 7 × 1.5 = 10.5 N

The force the ball exerts on the object = F₁ = m₁ × a₁ = 0.35 × a₁

By Newton's third law, F₁ = F₂, which gives;

m₁ × a₁ = m₂ × a₂

0.35 × a₁ = 10.5

a₁ = 10.5/0.35 = 30

The acceleration the ball gets = a₁ = 30 m/s².