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3 years ago

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Blocks A (mass 2.00 kg ) and B (mass 14.00 kg , to the right of A) move on a frictionless, horizontal surface. Initially, block

B is moving to the left at 0.500 m/s and block A is moving to the right at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is headon, so all motion before and after it is along a straight line. Let +x be the direction of the initial motion of A. Part A Find the maximum energy stored in the spring bumpers.

1 answer:

tia_tia [17]3 years ago

4 0

Answer:

7.72 Joules

Explanation:

Data:

mass of block A = 2.00 kg

mass of block B = 14.00 kg

velocity of block A = 2.00 m/s

velocity of block B = 0.500 m/s

The fundamental assumption is that there is elastic collision, therefore, the momentum and energy of the system are conserved.

Thus:

$U_{spring max} = KE_{total} - KE_{cm}\\ KE_{total} = \frac{1}{2}(mv_{1} ^{2} + mv_{2} ^{2} )\\ = 0.5 (3.00(2.00)^{2}+ 14.00 (0.500)^{2} \\ = 7.75 J$

$V_{cm} = \frac{3(2)-14(0.5)}{3+14} \\ = -0.0588m/s$

The kinetic energy is given by the following:

$KE_{cm} = \frac{1}{2}(m_{A} + m_{B})(v_{cm})^{2} \\ = \frac{1}{2}(3.00 + 14.00) (-0.0588)^{2} \\ = 0.03 J\\$

Therefore, the maximum energy is given by the following:

$U_{max} = 7.75 - 0.03\\ = 7.72 J$

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$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

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