You and a friend each hold a lump of wet clay. Each lump has a mass of 25 grams. You each toss your lump of clay into the air, where the lumps collide and stick together. Just before the impact, the velocity of one lump was < 4, 4, -3 > m/s, and the velocity of the other lump was < -2, 0, -7 > m/s. What was the the total momentum of the lumps just before the impact?
p(total) = kg·m/s.
What is the momentum of the stuck-together lump just after the collision?
p = kg·m/s.
What is the velocity of the stuck-together lump just after the collision?
v_f = m/s.

a) p(total) = <0.05, 0.1, 0.1 > kg m/s

b) p = <0.05, 0.1, 0.1 > kg.m/s

c) v_f = < 1, 2, 2 > m/s

Explanation:

a.)

Mass of each lump = 25 g = 0.025 kg

Velocity of lump 1 = < -2, 0, -7 > m/s

Momentum of lump 1 = Mass×Velocity

                                   = 0.025×< -2, 0, -7 >

                                   = < -0.05, 0, 0.175> kg m/s

Velocity of lump 2 = < 4, 4, -3 > m/s

Momentum of lump 2 = Mass×Velocity

                                    = 0.025×< 4, 4, -3 >

                                    = < 0.1, 0.1, -0.075> kg m/s

Total momentum before impact  =  < -0.05,  0,  0.175 > + < 0.1, 0.1, -0.075>

                                                      = < 0.05, 0.1, 0.1 > kg m/s

⇒p(total) = <0.05, 0.1, 0.1 > kg m/s

b)

As we know that,

By the law of conservation of linear momentum,

The total momentum will be the same before and after the collision.

⇒Momentum of the stuck together  after the collision = Total momentum of the lumps just before impact.

⇒ p = <0.05, 0.1, 0.1 > kg m/s

c)

Let the final velocity =  v_f

Total mass = 0.025 + 0.025 = 0.05 kg

As

Momentum = mass ×velocity

⇒ <0.05, 0.1, 0.1 > = 0.05 ×v_f

⇒ v_f = <0.05, 0.1, 0.1 > / 0.05

          = < 1, 2, 2 > m/s

⇒v_f = < 1, 2, 2 > m/s

its c i just checked it

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right