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A bullet of mass 30g is Firedfrom on of gun 9 kgiF thevelocity of bullet is 400m/s then the recoil velocity of the gun will be |
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Answer» :-RECOIL velocity of the gun is 1.33 m/s .Explanation :-We have :-→ Mass of the bullet (m) = 30 g = 0.03 kg→ Mass of the gun (M) = 9 kg→ Velocity of the bullet (m) = 400 m/sTo find :-→ Recoil velocity of the gun (V) .________________________________After the bullet is fired, the gun will recoil in the opposite direction (backwards) . So the value of recoil velocity will be -V m/s .According to "Law of CONSERVATION of Momentum", total momentum of the system will remain conserved. So, by using this concept, we have :-⇒ M(-V) + mv = 0⇒ -MV = -mv⇒ V = -mv/(-M)⇒ V = mv/MPutting VALUES in the obtained FORMULA, we get :-V = mv/M⇒ V = (0.03 × 400)/9⇒ V = 12/9⇒ V = 1.33 m/s |
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