Part 3 of 3 - Analyze The collision, for which figures and are before and after pictures, is perfectly inelastic, and momentum is conserved for the system of clay and block. We have m₁v₁ = (m₁ + m₂)v D 7.50 m- In the sliding process occurring between figures and, the original kinetic energy of the surface, block, and clay is equal to the increase in internal energy of the system due to friction. (m₁ + m₂)v₂² = FL Substituting the expression for fin terms of the total mass and friction coefficient, we have (m +m₂)v₂² = (m +m₂)gL Solving for 2 gives V₂ = √2μlg m/s. m (9.80 m/s²) Now from the momentum conservation equation, we have the following. = + M2 m₁ kg m/s) kg m/s. Submit Skip (you cannot come back)
Part 3 of 3 - Analyze The collision, for which figures and are before and after pictures, is perfectly inelastic, and momentum is conserved for the system of clay and block. We have m₁v₁ = (m₁ + m₂)v D 7.50 m- In the sliding process occurring between figures and, the original kinetic energy of the surface, block, and clay is equal to the increase in internal energy of the system due to friction. (m₁ + m₂)v₂² = FL Substituting the expression for fin terms of the total mass and friction coefficient, we have (m +m₂)v₂² = (m +m₂)gL Solving for 2 gives V₂ = √2μlg m/s. m (9.80 m/s²) Now from the momentum conservation equation, we have the following. = + M2 m₁ kg m/s) kg m/s. Submit Skip (you cannot come back)
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:Part 3 of 3 - Analyze
The collision, for which figures and
are before and after pictures, is perfectly inelastic, and momentum
is conserved for the system of clay and block. We have
m₁v₁ = (m₁ + m₂)v
D
7.50 m-
In the sliding process occurring between figures and, the original kinetic energy of the surface, block,
and clay is equal to the increase in internal energy of the system due to friction.
(m₁
+ m₂)v₂² = FL
Substituting the expression for fin terms of the total mass and friction coefficient, we have
(m
+m₂)v₂² = (m +m₂)gL
Solving for
2
gives
V₂ = √2μlg
m/s.
m
(9.80 m/s²)
Now from the momentum conservation equation, we have the following.
=
+ M2
m₁
kg
m/s)
kg
m/s.
Submit
Skip (you cannot come back)
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