An experiment is done to compare the initial speed of bullets fired from differe

Question

An experiment is done to compare the initial speed of bullets fired from different handguns: a 9 and a .44 caliber. The guns are fired into a 10- pendulum bob of length . Assume that the 9- bullet has a mass of 6 and the .44-caliber bullet has a mass of 12 . If the 9- bullet causes the pendulum to swing to a maximum angular displacement of 4.3 and the .44-caliber bullet causes a displacement of 10.1 , find the ratio of the initial speed of the 9- bullet to the speed of the .44-caliber bullet, .

Explanation / Answer

For 9mm bullet:- Let the initial velocity of bullet is U9 m/s, vertical height gained by pendulum (h) = L - LcosA* = L(1-cos4.3*) = 2.8 x 10^-3 x L Thus by the law of energy conservation:- =>KE = PE =>1/2(M+m) x v^2 = (M+m) x gh =>v^2 = 2gh =>v = sqrt[2 x 9.8 x 2.8 x 10^-3 x L] By the law of momentum conservation:- m1u1+m2u2 = (m1+m2) x v =>6 x 10^-3 x U9 + 0 = (10+6 x 10^-3) x sqrt[2 x 9.8 x 2.8 x 10^-3 x L] =>sqrt[2 x 9.8 x 2.8 x 10^-3 x L] = 6 x 10^-4 x U9 =>[2 x 9.8 x 2.8 x 10^-3 x L] = 36 x 10^-8 x (U9)^2 =>(U9)^2 = 1.52 x 10^5 x L ---------------------(i) For the 0.44 caliber bullet:- Let the initial velocity of bullet is U0.44 m/s, vertical height gained by pendulum (h) = L - LcosA* = L(1-cos10.1*) = 15.50 x 10^-3 x L Thus by the law of energy conservation:- =>KE = PE =>1/2(M+m) x v^2 = (M+m) x gh =>v^2 = 2gh =>v = sqrt[2 x 9.8 x 15.50 x 10^-3 x L] By the law of momentum conservation:- m1u1+m2u2 = (m1+m2) x v =>12 x 10^-3 x U0.44 + 0 = (10+12 x 10^-3) x sqrt[2 x 9.8 x 15.50 x 10^-3 x L] =>sqrt[2 x 9.8 x 15.50 x 10^-3 x L] = 11.99 x 10^-4 x U0.44 =>[2 x 9.8 x 15.50 x 10^-3 x L] = 143.76 x 10^-8 x (U0.44)^2 =>(U0.44)^2 = 2.11 x 10^5 x L ---------------------(ii) =>By (i)/(ii) :- =>(U9)^2/(U0.44)^2 = 1.52/2.11 = 0.72 =>U9/U0.44 = sqrt0.72 = 0.85

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