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A man standing on very slick ice fires a rifle horizontally. The mass of the man together with the rifle is 70 kg , and the mass of the bullet is 10 g . If the bullet leaves the muzzle at a speed of 500 m/s , what is the final speed of the man?

 Feb 17, 2016

Best Answer 

 #3
avatar+1312 
+5

This is called conservation of momentum

Mass * velocity  =   momentum

Solve it like this

 

m1 * v1 = m2 * v2

0.010kg x 500m/s = 70kg * v2

(0.010*500)/70 =0.0714

 

The final speed of the man is 0.0714 m/s

 Feb 17, 2016
 #1
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Zero b***h

 Feb 17, 2016
 #2
avatar+333 
0

we must first figure out the force of the bullet with

F=ma so 10g*500  =5000N

So the bullet will produce a horizontal force of 5000N

so 5000N at the speed of 500/ms

the final speed is 5000/500=10/ms(^-2) in horizontal force

Sorry if i made some mistakes, i'm not amazing at physics

 Feb 17, 2016
 #3
avatar+1312 
+5
Best Answer

This is called conservation of momentum

Mass * velocity  =   momentum

Solve it like this

 

m1 * v1 = m2 * v2

0.010kg x 500m/s = 70kg * v2

(0.010*500)/70 =0.0714

 

The final speed of the man is 0.0714 m/s

Dragonlance Feb 17, 2016
 #4
avatar+105509 
0

Physics is neat!  Thanks Dragonlance.

 Feb 17, 2016

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