Question

a shell of mass 10 kg is fired with a velocity of 80 m/s making an angle 30 with the vertical. Find its momentum when the shell reaches the highest position ​

Answer 1

Answer:

400 kgms^-1

Explanation:

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Answer 2

Concept

• If the net force on a body is zero in a certain direction, the body remains at rest or moves uniformly in that direction.

• The momentum of a body is given by the product of mass and velocity i.e. p = mv.

• At the highest position, the projectile momentarily stops in the vertical direction.

Given

• The mass of the shell is 10 kg.
• The initial velocity of the shell is 80 m/s.
• The initial velocity makes 30° angle with the vertical.

Find

The momentum of the shell when it reaches the highest position.

Solution

Horizontal velocity of the shell

In the horizontal direction, there is no force acting on the shell. So, the shell will moves uniformly in the horizontal direction, the horizontal velocity of the shell being equal to the initial horizontal velocity.

Horizontal velocity, vₓ = v$\mathbf{\hat i}$ sinθ = (80$\mathbf{\hat i}$ sin30°)m/s = 40$\mathbf{\hat i}$ m/s.

Velocity of the shell at the highest position

At the highest position, the shell momentarily stops in the vertical direction. So, the vertical velocity at the highest position is zero.

So, the net velocity of the particle is the horizontal velocity.

Velocity of the particle at the highest position, $\mathbf{v}_h$ = $\mathbf{v}_x$ = 40$\mathbf{\hat i}$ m/s.

The momentum of the shell at the point of maximum height

The momentum of the shell, $\mathbf{p}_h$ = $m\mathbf{v}_h$ = 10 × 40$\mathbf{\hat i}$ = 400$\mathbf{\hat i}$ kg·m/s.

The momentum of the shell at the highest position is 400$\mathbf{\hat i}$ kg·m/s.

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