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At the top of a trajectory, a mortar shell explodes into two fragments; a 2.5 kg piece mo south west with a horizontal velocity of 100 m/s while 3.5 kg piece moves north horizontal velocity of 70 m/s. What was the horizontal direction of the cell just before explosion​

User Dmytro Maslenko
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To find the horizontal direction of the shell just before the explosion, you can use the principle of conservation of momentum. This principle states that the total momentum of a closed system (one that is isolated from its surroundings) remains constant, provided that there are no external forces acting on the system.

In this case, the shell and the two fragments form a closed system, since no external forces are acting on them. Before the explosion, the shell has a certain momentum, and after the explosion, the two fragments have a combined momentum that is equal to the initial momentum of the shell.

To find the initial momentum of the shell, you can use the formula:

p = m * v

where p is the momentum, m is the mass of the object, and v is the velocity.

The initial momentum of the shell is equal to the combined momentum of the two fragments after the explosion:

p = (2.5 kg * 100 m/s) + (3.5 kg * 70 m/s) = 250 kg m/s + 245 kg m/s = 495 kg m/s

Since the mass of the shell is not given, it's not possible to find the initial velocity of the shell. However, you can find the direction of the initial velocity by looking at the direction of the final velocities of the two fragments.

The direction of the final velocity of the 2.5 kg fragment is south west, and the direction of the final velocity of the 3.5 kg fragment is north. This means that the direction of the initial velocity of the shell was likely somewhere between these two directions.

It's not possible to determine the exact direction of the initial velocity of the shell without more information, but based on the given information, it was likely somewhere between south west and north.

User Conchi
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