A bullet of mass 20 g, travelling at a speed of 350 ms−1 , strikes a steel plate at an angle of 30º with the plane of the plate. It ricochets off at the same angle, at a speed of 320 ms−1 . What is the magnitude of the impulse that the steel plate gives to the projectile? If the collision with the plate takes place over a time ∆t = 10−3 s, what is the average force exerted by the plate on the bullet?
kvnmurty:
speeds are not same... 350 and 320..
did i write in the answer that speed are same?
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See the diagram.
m = 0.020 kg. Δt = 0.001 sec.
u = 350 m/s. Ф1 = 30°. v = 320 m/s. Ф2 = Ф1 = 30°.
If u cos Ф1 = v cos Ф2, the component of speed (& momentum) along the plane of the plane remains same afer the collision. Then the force exerted and the change in momentum (IMPULSE) will be only in the direction perpendicular to the plate.
But here it is not the same case as v ≠ u...
Change in momentum horizontally = m v cos Ф2 - m u cos Ф1
= 0.020 cos30 *(320 - 350) = - 0.5196 kg-m/s
Change in momentum vertically: m v sinФ2 - (- m u sinФ1)
= 0.020 * 1/2 * (320+350) = 6.700 kg-m/s
Magnitude of change in momentum
= √[0.5196² + 6.700²] = 6.720 kg-m/s
Average force (magnitude) = 6.720 /0.001 = 6720 N.
m = 0.020 kg. Δt = 0.001 sec.
u = 350 m/s. Ф1 = 30°. v = 320 m/s. Ф2 = Ф1 = 30°.
If u cos Ф1 = v cos Ф2, the component of speed (& momentum) along the plane of the plane remains same afer the collision. Then the force exerted and the change in momentum (IMPULSE) will be only in the direction perpendicular to the plate.
But here it is not the same case as v ≠ u...
Change in momentum horizontally = m v cos Ф2 - m u cos Ф1
= 0.020 cos30 *(320 - 350) = - 0.5196 kg-m/s
Change in momentum vertically: m v sinФ2 - (- m u sinФ1)
= 0.020 * 1/2 * (320+350) = 6.700 kg-m/s
Magnitude of change in momentum
= √[0.5196² + 6.700²] = 6.720 kg-m/s
Average force (magnitude) = 6.720 /0.001 = 6720 N.
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