A bullet of mass 20g travelling at speed of 350m per second strike a steel plate at an angle of 30degree with the plane of plate it ricochets off at the same angle at a speed of 320m per second what is the magnitude of impulse that the steel plates gives to the projectile if the collision with with the plates take places over a time dt=10 raise to power -3s.what is average force exerted by the plateson the bullet
To calculate the magnitude of impulse and average force exerted by the steel plate on the bullet, we can start by using the formula:
Impulse (J) = Change in momentum (Δp)
1. Find the initial momentum of the bullet:
Momentum (p) = mass (m) * velocity (v)
mass = 20g = 0.02kg
velocity = 350m/s
Initial momentum (p1) = (0.02kg) * (350m/s)
= 7 kg·m/s
2. Find the final momentum of the bullet:
Final velocity = 320m/s
Final momentum (p2) = (0.02kg) * (320m/s)
= 6.4 kg·m/s
3. Calculate the change in momentum:
Δp = p2 - p1
= 6.4 kg·m/s - 7 kg·m/s
= -0.6 kg·m/s (The negative sign indicates a reversal of direction.)
4. Calculate the magnitude of impulse:
|Δp| = |-0.6 kg·m/s|
= 0.6 kg·m/s
So, the magnitude of the impulse that the steel plate gives to the projectile is 0.6 kg·m/s.
To determine the average force exerted by the steel plate on the bullet, we can use the formula:
Average force (F_avg) = Δp / Δt
Given that Δt = 10^-3 s, substitute the values and calculate:
F_avg = 0.6 kg·m/s / (10^-3 s)
= 0.6 N
Therefore, the average force exerted by the steel plate on the bullet is 0.6 Newtons.