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a bullet of mass 25 g is fired horizontally into a sand-filled box which is suspended by long strings from the ceiling. The combined mass of the bullet, box, and sand is 10 kg. After impact the box and bullet reach a max height of 0.1m. find the max velocity of the box

PE = mgh

on impact, this was all KE = 1/2 mv^2
So solve for v.

To find the maximum velocity of the box after the impact, we can apply the law of conservation of momentum and the law of conservation of mechanical energy.

Let's start by analyzing the situation using the conservation of momentum:

The initial momentum of the bullet is given by:
𝑝𝑖 = π‘šπ‘ Γ— 𝑣𝑏
Where π‘šπ‘ is the mass of the bullet (25 g = 0.025 kg), and 𝑣𝑏 is the initial velocity of the bullet before impact.

Since the bullet is fired horizontally, the vertical component of its initial velocity is zero. Therefore, 𝑣𝑏 = 0.

The total momentum before the impact is equal to the total momentum after the impact since no external forces act on the system. Since the box and the sand are at rest initially, their momentum is zero.

After the impact, both the bullet and the sand-filled box will move vertically, and their combined momentum is given by:
𝑝𝑓 = (π‘šπ‘ + π‘šπ‘  + π‘šπ‘) Γ— 𝑣𝑓
Where π‘šπ‘  is the mass of the sand.

But since the box and bullet reach a maximum height, their final velocity in the upward direction becomes zero, 𝑣𝑓 = 0. Therefore, 𝑝𝑓 = 0.

Equating the initial and final momentum, we have:
π‘šπ‘ Γ— 𝑣𝑏 = (π‘šπ‘ + π‘šπ‘  + π‘šπ‘) Γ— 𝑣𝑓
0.025 kg Γ— 0 = (0.025 kg + π‘šπ‘  + 0.025 kg) Γ— 0
0 = 0.05 kg + π‘šπ‘  + 0.05 kg

Simplifying the equation, we find:
0.05 kg = -2π‘šπ‘ 

Since the mass of the sand cannot be negative, -2π‘šπ‘  = 0.05 kg implies that π‘šπ‘  = 0.025 kg.

Now that we know the mass of the sand, we can calculate the kinetic energy of the system before the impact using the conservation of mechanical energy:

The initial kinetic energy is given by:
𝐾𝑖 = 0.5 Γ— π‘šπ‘ Γ— 𝑣𝑏^2
Where 𝑣𝑏 = 0.

The final kinetic energy is given by:
𝐾𝑓 = 0.5 Γ— (π‘šπ‘ + π‘šπ‘  + π‘šπ‘) Γ— 𝑣𝑓^2
Where 𝑣𝑓 = 0.

Equating the initial and final kinetic energy, we have:
0.5 Γ— 0.025 kg Γ— 0 = 0.5 Γ— (0.025 kg + 0.025 kg + 0.025 kg) Γ— 0

Therefore, the initial kinetic energy 𝐾𝑖 = 0 J.

Now, we can find the maximum velocity of the box using the principle of conservation of mechanical energy:

The initial potential energy is zero since the box is at rest before the impact.

The final potential energy is given by the equation:
𝑃𝑓 = (π‘šπ‘ + π‘šπ‘  + π‘šπ‘) Γ— 𝑔 Γ— β„Žπ‘“
Where β„Žπ‘“ is the maximum height reached by the box (0.1 m), and 𝑔 is the acceleration due to gravity (approximately 9.8 m/s^2).

Equating the initial and final potential energy, we have:
0 = (0.025 kg + 0.025 kg + 0.025 kg) Γ— 9.8 m/s^2 Γ— 0.1 m

Simplifying the equation, we find:
0 = 0.075 kg Γ— 9.8 m/s^2 Γ— 0.1 m
0 = 0.0735 N

Since the potential energy is zero, the maximum velocity of the box is also zero, because there is no kinetic energy.

Therefore, the maximum velocity of the box after the impact is 0 m/s.