A1-kg chunk of putty moving at 1 m/s collides with and sticks to a 5-kg bowling ball initially at rest. the bowling ball and putty then move with a momentum of a. 0 kg m/s. b. 1 kg m/s. c. 2 kg m/s. d. 5 kg m/s. e. more than 5 kg m/s.

b) 1kg m/s

Explanation:

∑p = ∑mv = 1kg * 1m/s + 5kg * 0 m/s = 1 kg m/s

The bowling ball and putty move with a momentum of 1kg m/s ( option B)

Explanation:

For this situation, we use the Law of Conservation of momentum'

This law describes that the sum of the initial momentums equals the sum of the final momentum of colliding bodies, and is given by following equation:

p = m*v

The momentum of a particle or an object is the product of two quantities, the particle's mass (represented by the letter m) and its velocity (v). The momentum of the particle/ object itself is shown by the letter p.

Pbefore = Pafter

m1u1 + m2u2 = m1v1 + m2v2

⇒ with u1 and u2 = initial velocities

⇒ With V1 and V2 = the final velocities V1 and V2 → those are equal to V

So we have:

m1u1 +m2u2 = (m1+m2)V

(m1+m2)V = the final momentum of the bowling ball + putty (Pafter)

1*1 + 5*0 = (m1 + m2)V = Pafter

Pafter = 1kg * m/s

The bowling ball and putty move with a momentum of 1kg m/s

explanation:

v=uf/(u-f)

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