Question

Two carts are connected by a loaded spring on a horizontal, frictionless surface. The spring is released and the carts push away from each other. Cart 1 has mass M and Cart 2 has mass M/3.

a) Is the momentum of Cart 1 conserved?

Yes

No

It depends on M

b) Is the momentum of Cart 2 conserved?

Yes

No

It depends on M

c) Is the total momentum of Carts 1 and 2 conserved?

Yes

No

It depends on M

d) Which cart ends up moving faster?

Cart 1

Cart 2

They move at the same speed

e) If M = 6 kg and Cart 1 moves with a speed of 16 m/s, what is the speed of Cart 2?

0 m/s

4.0 m/s

5.3 m/s

16 m/s

48 m/s

64 m/s

Answer 1

a) yes

b) no

c) yes

d)Cart 2 with mass
`(M)/(3)` is expected to be more faster

e) u₂ = 48 m/s

Explanation:

a) the all out linear momentum of an arrangement of particles of Cart 1 not followed up on by external forces is constant.

b) the linear momentum of Cart 2 will be acted upon by external force by Cart 1 with mass M, thereby it's variable and the momentum is not conserved

c) yes, the momentum is conserved because no external force acted upon it and both Carts share the same velocity after the reaction

note: m₁u₁ + m₂u₂ = (m₁ + m₂)v

d) Cart 2 with mass
`(M)/(3)` will be faster than Cart 1 because Cart 2 is three times lighter than Cart 1.

e) Given

m₁= M

u₁ = 16m/s

m₂ =
`(M)/(3)`

u₂ = ?

from law of conservation of momentum

m₁u₁= m₂u₂

M× 16 =
`(M)/(3)` × u₂(multiply both sides by 3)

therefore, u₂ =
`(3(M .16))/(M)` ("." means multiplication)

∴u₂ = 3×16 = 48 m/s