A bullet of mass 10 grams travelling horizontally with a vertical 150 M per second of Strike a stationary wooden block and coming to rest in 0.003 seconds calculate the distance of penetration of the bullet into the wooden block also calculate the also calculate the magnitude of the fore exerted by the wooden block on the bullet. ( it is a 3 mark question answer it fast )
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fSolution:-
bullet mass (m) = 10grams = 10/1000kg
ve
u = 150m/s , v = 0 , t= 0.003 secound
aplied rest = a = (v-u)/t = (0-150)/0.003 = -150/0.003
=- 150000/3 = -50000m/s^2
= block distance = s
then
s= ut + 1/2at^2 = 150×0.003 + 1/2×50000×(0.003)^2
s = 0.45 +0.225 = 0.270mtr
by f= ma
f = 10/1000×50000
[ f = 10×50 = 500newton] ans
bullet mass (m) = 10grams = 10/1000kg
ve
u = 150m/s , v = 0 , t= 0.003 secound
aplied rest = a = (v-u)/t = (0-150)/0.003 = -150/0.003
=- 150000/3 = -50000m/s^2
= block distance = s
then
s= ut + 1/2at^2 = 150×0.003 + 1/2×50000×(0.003)^2
s = 0.45 +0.225 = 0.270mtr
by f= ma
f = 10/1000×50000
[ f = 10×50 = 500newton] ans
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