A26-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.

  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

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

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explanation:

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.