A bullet of mass 20g, travelling at a speed of 350m/s, strikes a steel plate at an angle of 30° with the plane of the plate. It ricochet off at the same angle at a speed of 320m/s. What is the magnitude of the impulse that the steel plate given to the projectile? If the collision with the plate take place over a time ^t=10m/s , what is the average force exerted by the plate on the bullet?
force*time=changemomentum
Because the force is a vector, we how shift to vectors.
In the direction normal to the plate:
forceN*time=mass*(Vnf-Vni)
forceN=.020 (350sin30-(-320sin30))/.01
solve that, that is the average force normal to the plate, in the direction up from the plate.
In the direction Parallel to the plate:
forceP=.020(350cos30-320cos30)/.010
solve that.
Then average force= sqrt(Fn^2+Fp^2) in magnitude, to solve the direction, use your trig relationships. Fn is much greater than Fp
To find the magnitude of the impulse that the steel plate gives to the projectile, we can use the principle of conservation of momentum.
The momentum of an object is given by the product of its mass and velocity. The momentum before the collision is equal to the momentum after the collision, provided no external forces are acting on the system.
Given:
Mass of the bullet, m = 20g = 0.02kg
Initial velocity, u = 350m/s
Final velocity, v = 320m/s
First, we need to find the initial and final momenta of the bullet:
Initial momentum, P1 = mass × initial velocity = 0.02kg × 350m/s
Final momentum, P2 = mass × final velocity = 0.02kg × 320m/s
The change in momentum or impulse is given by:
Impulse = final momentum - initial momentum = P2 - P1
Now, to find the average force exerted by the plate on the bullet, we can use the equation:
Average force = impulse / time
Given:
Time of collision, Δt = 10ms = 0.01s
Hence, to find the solution, we need to calculate the impulse and average force using the above formulas.