Question

A metal sphere with a mass of 0.026 kg rolls along a friction-less surface at 3.7 m/s and strikes a stationary sphefe having a mass of 0.050 kg. The first sphere stops completely. The spheres are in contact for 0.007s before the second sphere is shot off down the track. What is the force applied to the second ball?

Answer 1

The force applied to the second ball by the first ball is 6.734 × 10^-4 N.

What is impulse of force?

The impulse of force is defined as the sum of the average force and the duration it is applied.

If the mass of the item remains constant, the impulse of force equals the change in momentum of the object.

Given that:  mass of a metal sphere: m = 0.026 kg.

Initial speed of the sphere: u = 3.7 m/s.

When the sphere stops completely, its change in momentum = mu - 0

= 0.026×3.7 N-s.

= 0.0962 N-s.

As the spheres are in contact for 0.007s before the second sphere is shot off down the track, the force applied  to the second ball =

change in momentum of 1st ball × time of contact

= 0.0962 × 0.007 N

= 0.0006734 N

= 6.734 × 10^-4 N.

Hence,  the force applied  to the second ball is  6.734 × 10^-4 N.

Learn more about impulse force here:

brainly.com/question/29787329

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