A 26-g steel-jacketed bullet is fired with a velocity of 630 m/s toward a steel plate and ricochets along path CD with a velocity 500 m/s. Knowing that the bullet leaves a 50-mm scratch on the surface of the plate and assuming that it has an average speed of 600 m/s while in contact with the plate, determine the magnitude and direction of the impulsive force exerted by the plate on the bullet.

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moya1316 moya1316 · Answer 1 · 2019-12-23T09:43:41+01:00

Answer:

F = - 3.53 10⁵ N

Explanation:

This problem must be solved using the relationship between momentum and the amount of movement.

I = F t = Δp

To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio

v = d / t

t = d / v

Reduce SI system

m = 26 g ( 1 kg/1000g) = 26 10⁻³ kg

d = 50 mm ( 1m/ 1000 mm) = 50 10⁻³ m

Let's calculate

t = 50 10⁻³ / 600

t = 8.33 10⁻⁵ s

With this value we use the momentum and momentum relationship

F t = m v - m v₀

As the bullet bounces the speed sign after the crash is negative

F = m (v-vo) / t

F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵

F = - 3.53 10⁵ N

The negative sign indicates that the force is exerted against the bullet