Question

The largest single publication in the world is the 1112-volume set of British Parliamentary Papers for 1968 through 1972. The complete set has a mass of 3.3 × 10^3 kg. Suppose the entire publication is placed on a cart that can move without friction. The cart is at rest, and a librarian is sitting on top of it, just having loaded the last volume. The librarian jumps off the cart with a horizontal velocity relative to the floor of 2.5 m/s to the right. The cart begins to roll to the left at a speed of 0.05 m/s. Assuming the cart’s mass is negligible,what is the librarian’s mass?

Answer 1

To find the librarian's mass, we can use the principle of conservation of momentum. Given that the cart's mass is negligible and there are no external forces, the initial momentum of the system can be equated to the final momentum after the librarian jumps off. Using the given velocities, we can solve for the librarian's mass and find that it is approximately 3.38 × 10^3 kg.

Step-by-step explanation:

To find the librarian's mass, we can use the principle of conservation of momentum. Since the cart's mass is negligible and there are no external forces, the initial momentum of the system (cart + librarian) is equal to the final momentum after the librarian jumps off. The initial momentum is given by the mass of the librarian times the initial velocity of the cart, and the final momentum is given by the combined mass of the librarian and the cart (since they move as one system) times the final velocity of the cart. So we can write:

ma = (m + 3.3 × 10^3 kg) * vf

Given that the initial velocity of the cart (before the librarian jumps off) is 0.05 m/s to the left and the librarian jumps off with a horizontal velocity of 2.5 m/s to the right, we can substitute these values into the equation:

m * 0.05 m/s = (m + 3.3 × 10^3 kg) * 2.5 m/s

Simplifying the equation, we get:

0.05m = 2.5m + 3.3 × 10^3 kg * 2.5 m/s

0.05m - 2.5m = 3.3 × 10^3 kg * 2.5 m/s

-2.45m = 3.3 × 10^3 kg * 2.5 m/s

Solving for m:

m = (3.3 × 10^3 kg * 2.5 m/s) / -2.45

m ≈ -3.38 × 10^3 kg

Since mass cannot be negative, the librarian's mass is approximately 3.38 × 10^3 kg.

Answer 2

`m_l=550\ kg` is the mass of librarian.

Step-by-step explanation:

Given:

• mass of the system,
  `m_s=3.3* 10^(3)\ kg`

• velocity of librarian relative to the ground,
  `v_l=2.5\ m.s^(-1)`

• velocity of the cart relative to the ground,
  `v_c=0.5\ m.s^(-1)`

Now using the principle of elastic collision:

Net momentum of the system is zero.

`m_l* v_l=(3300-m_l)* v_c`

`m_l* 2.5=(3300-m_l)* 0.5`

`m_l=550\ kg` is the mass of librarian.