Asked by Jitender
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?
Answers
Answered by
bobpursley
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
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
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