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3 years ago

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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, w

here 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.

1 answer:

solniwko [45]3 years ago

7 0

Answer:

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

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