A 0.020-kg bullet traveling at a speed of embeds in a 1.0-kg wooden block resting on a horizontal surface. The block slides horizontally 4.0 m on a surface before stopping. Determine the coefficient of kinetic friction between the block and surface.
A 0.020-kg bullet traveling at a speed of embeds in a 1.0-kg wooden block resting on a horizontal surface. The block slides horizontally 4.0 m on a surface before stopping. Determine the coefficient of kinetic friction between the block and surface.
I know the solution is to first use 1) M1V1 + 0 = (M1+M2) Vf1 to get Vf1 = 5.9m/s
Then use 2) net force =
However I'm confused why I couldn't use p + J = p for the second part after I got Vf1. In which case I would have (m1+m2)Vf1 + Ff*d = 0,
(m1+m2)Vf1 = (m1+m2)g *Uk *d
Uk = Vf1/gd
When I use this I get 0.15, which is the wrong answer. The right answer is 0.44. Can you help me identify the mistake in my logic? or if this is a way i could use but I'm just missing pieces? Thank you so much.
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Solve this using momentum and not conservation of energy:
A 0.020-kg bullet traveling at a speed of embeds in a 1.0-kg wooden block resting on a horizontal surface. The block slides horizontally 4.0 m on a surface before stopping. Determine the coefficient of kinetic friction between the block and surface.
I know the solution is to first use 1) M1V1 + 0 = (M1+M2) Vf1 to get Vf1 = 5.9m/s
Then use 2) net force =
However I'm confused why I couldn't use p + J = p for the second part after I got Vf1. In which case I would have (m1+m2)Vf1 + Ff*d = 0,
(m1+m2)Vf1 = (m1+m2)g *Uk *d
Uk = Vf1/gd
When I use this I get 0.15, which is the wrong answer. The right answer is 0.44. Can you help me identify the mistake in my logic? or if this is a way I could use but I'm just missing pieces? Thank you so much.
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