Question

First ANSWER = BRAINLIESTquestion 3 onwards

Answer 1

3. Conservation of momentum for inelastic collisions (the trolleys collide and stick together):

v = (m₁v₁+m₂v₂)/(m₁+m₂)

Given values:

m₁ = m₂ = 1kg

v₁ = 5m/s

v₂ = 0m/s

Substitute the terms in the equation with the given values and solve for v:

v = (1×5+1×0)/(1+1)

v = 2.5m/s

4. Apply the same equation for inelastic collisions (the skaters hold on together after the collision):

v = (m₁v₁+m₂v₂)/(m₁+m₂)

m₁ = 60kg (moving skater's mass)

m₂ = 30kg (stationary skater's mass)

v₁ = 6m/s (moving skater's speed)

v₂ = 0m/s (stationary skater is at rest)

Substitute the values and solve for v:

v = (60×6+30×0)/(60+30)

v = 4m/s

5.

a) Since the car and lorry are moving as one object after the collision, multiply the sum of their masses by their velocity to get their momentum:

p = v×∑m

Given values:

m₁ = 1000kg (sports car)

m₂ = 3000kg (lorry)

v = 25 m/s

Plug in the values and solve for p:

p = 25×(1000+3000)

p = 100000$\frac{kg×m}{s}$

b) The conservation of momentum guarantees that the total momentum of the objects before and after the collision will be equal. Therefore the total momentum before the collision is the same as the value obtained in the previous question:

p = 100000$\frac{kg×m}{s}$

c) Assume the lorry was at rest before the collision.

Recall the general equation for conservation of momentum:

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

Given values:

m₁ = 1000kg (sports car)

m₂ = 3000kg (lorry)

v₂ = 0m/s (lorry is initially at rest)

v₁' = v₂' = 25m/s (car and lorry stick together after collision, moving at the same speed)

Plug in the given values and solve for v₁ (car's speed before the collision):

1000v₁ + 3000×0 = 1000×25 + 3000×25

v₁ = 100m/s