A riffle man, who together with his rifle has a mass of 200 kg, stands on a smooth surface and fires 10 shots horizontally. Each bullet has a mass of 20 g and muzzle velocity of 600 m/s. the velocity which rifle man attains after firing 10 shots is What?
Answers
Answer:
a. Let m1 and m2 be the masses of bullet and the rifleman, v1 and v2 their respective velocities after the first shot.
Initially, the rifleman and bullet are at rest, therefore initial momentum of the system =0.
As external force is zero, momentum of the system is constant i.e., initial momentum = final momentum 0=m1v1=m2v2
or v2=−m1v1m2=−(10×10−3kg)(800m/s)100kg=−0.08m/s
As mass of the bullet is negligible as compared to (rifle + man),
∴ Velocity acquired after 10 shots =10v2= 0.8 m / s
i.e., the velocity of rifleman is 0.8 m / secin a direction opposite to that of the bullet.
b. The momentum is acquired by the rifleman is
P=m2v2=100×0.8kgm/s=80kgm/s
This momentum is acquired in 10s, therefore the average force exeed =△p△t=8010= 8 N
c. Kinetic energy of the rifleman,
E1=12m2v22=12×100(0.8)2= 32 J
Kinetic energyy of 10 bullet
E2=10[12×10×10−3]=(800)2=32000J
Kinetic energy of bullet is 1000 times the energy of rifleman.
Explanation:
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Answer: Given, Mass,M=100kg,m=10g,t=5s
By the law of momentum conservation:-
=>P(iniitial)=P(final)=>mu
1
+Mu
2
=mv
1
+Mv
2
=>0+0=nmu+(M−nm)v
=>10×10×10
−3
×800+(100−10×10×10
−3
)×v=>v=−0.80m/s
{−ve just indicating the direction is opposite to the direction of bullets}
Now, By F=
Δt
ΔP
=>F=
5
10×10×10
−3
×800
=16N
Explanation: PLEASE MAKE ME THE BRAINLEIST