nbsp-Q28 answer fast-Q28- A bullet of mass 40 g is fired from a gun of mass 10 kg- If the velocity of bullet is 400 m-s- Then the recoil velocity of the gun will be nbsp-1- 1-6 m-s in the direction of bullet-2- 1-6 m-s opposite to the direction of bullet-3- 1-8 m-s in the direction of bullet-4- 1-8 cm-s opposite to the direction of bullet- nbsp-

Dear #160;student,by #160;law #160;of #160;conservation #160;of #160;momentuminitial #160;momentum=final #160;momentum0=m1 #215;v1+m2 #215;r ...

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Byju's Answer

Standard XII

Physics

Instantaneous Force and Impulse

nbsp; Q28 an...

Question

Q28 answer fast

Q28. A bullet of mass 40 g is fired from a gun of mass 10 kg. If the velocity of bullet is 400 m/s. Then the recoil velocity of the gun will be

(1) 1.6 m/s in the direction of bullet

(2) 1.6 m/s opposite to the direction of bullet

(3) 1.8 m/s in the direction of bullet

(4) 1.8 cm/s opposite to the direction of bullet

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Solution

$D e a r s t u d e n t, b y l a w o f c o n s e r v a t i o n o f m o m e n t u m i n i t i a l m o m e n t u m = f i n a l m o m e n t u m 0 = m_{1} \times v_{1} + m_{2} \times r e c o i l v e l o c i t y r e c o i l v e l o c i t y = - \frac{m_{1}}{m_{2}} v_{1} = - \frac{40 \times 400}{10000} = - 1.6 m / s s o t h e v e l o c i t y w i l l b e 1.6 m / s o p p o s i t e t o t h e d i r e c t i o n o f b u l l e t R e g a r d s$

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Dear student,by law of conservation of momentuminitial momentum=final momentum0=m1×v1+m2×recoil velocityrecoil velocity=-m1m2v1=-40×40010000=-1.6m/sso the velocity will be 1.6m/s opposite to the direction of bulletRegards