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?
+1316 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
+333 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
+1316 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
