Question

a rifle bullet with a muzzle speed of is fired directly into a special dense material that stops the bullet in 25.0 cm. assuming the bullet’s deceleration to be constant, what is its magnitude?

Answer 1

The deceleration of a bullet with a muzzle velocity of 350 m/s, which stops in 25.0 cm, is calculated to be 490000 m/s².

Step-by-step explanation:

The task involves calculating the deceleration magnitude of a bullet that comes to rest after penetrating a dense material. Given the muzzle velocity (initial velocity) and the stopping distance, we can employ kinematic equations to find the deceleration. Assuming constant deceleration, we use the equation ² = ₀² + 2ad, where v is the final velocity (0 m/s since the bullet stops), ₀ is the initial velocity, a is the deceleration, and d is the stopping distance.

Given: ₀ = 350 m/s and d = 0.25 m, and knowing that v = 0 m/s because the bullet stops, we can rearrange the equation to solve for a: a = (² - ₀²) / (2d).

Plugging the values in, we get: a = (0 - (350 m/s)²) / (2 × 0.25 m), which leads to a = -490000 m/s². The negative sign indicates the deceleration is in the opposite direction of the bullet's movement.

Deceleration

of the bullet is 490000 m/s².