Question

A girl of mass m1=60.0 kilograms springs from a trampoline with an initial upward velocity of vi=8.00 meters per second. At height h=2.00 meters above the trampoline, the girl grabs a box of mass m2=15.0 kilograms. (Figure 1) For this problem, use g=9.80 meters per second per second for the magnitude of the acceleration due to gravity.

A girl of mass m1=60.0 kilograms springs from a trampoline with an initial upward velocity of vi=8.00meters per second. At height h=2.00 meters above the trampoline, the girl grabs a box of mass m2=15.0kilograms. (Figure 1)

For this problem, use g=9.80 meters per second per second for the magnitude of the acceleration due to gravity.

A) What is the speed vbefore of the girl immediately before she grabs the box?

B) What is the speed vafter of the girl immediately after she grabs the box?

C) What is the maximum height hmax that the girl (with box) reaches? Measure hmax with respect to the top of the trampoline.

Answer 1

Concepts and reason

The concept required to solve the given problem is the law of conservation of linear momentum and law of conservation of energy.

Calculate the speed of the girl before grabbing the box and the maximum height reached by the girl with the help of law of conservation of energy.

For the second part use the law of conservation of momentum to calculate the speed of the girl after grabbing the box.

Fundamentals

Law of conservation of linear momentum: It states that when no external force acts on a body the momentum remains conserved. Mathematically the statement may be expressed as,

${p_{\rm{i}}} = {p_{\rm{f}}}$

Here, ${p_{\rm{i}}}$ and ${p_{\rm{f}}}$ is the initial and final momentum of the body into consideration.

Momentum is given by,

$p = mv$

Here, $m$ is the mass and $v$ is the velocity of the body.

Law of conservation of energy: It states that the energy can neither be created nor destroyed but can only be changed from one form to another. Mathematically it may be written as,

$U + K = {\rm{ constant}}$

Or,

$\begin{array}{c}\\{U_{\rm{f}}} + {K_{\rm{f}}} = {U_{\rm{i}}} + {K_{\rm{i}}}\\\\{U_{\rm{i}}} - {U_{\rm{f}}} = {K_{\rm{f}}} - {K_{\rm{i}}}\\\\\left| {\Delta U} \right| = \left| {\Delta K} \right|\\\end{array}$

Here, ${U_{\rm{f}}}{\rm{ and }}{U_{\rm{i}}}$ are the final and initial potential energy respectively and ${K_{\rm{f}}}{\rm{ and }}{K_{\rm{i}}}$ are the initial and final kinetic energy respectively.

(A)

Apply the law of conservation of energy, which says the gain in potential energy is equal to the loss in kinetic energy.

${U_{\rm{i}}} - {U_{\rm{f}}} = {K_{\rm{f}}} - {K_{\rm{i}}}$

Here, ${U_{\rm{i}}}$ and ${U_{\rm{f}}}$ are the initial and final potential energy respectively and ${K_{\rm{i}}}{\rm{ }}$and ${K_{\rm{f}}}$ are the initial and final kinetic energy respectively.

The kinetic energy is given by,

$K = \frac{1}{2}m{v^2}$

Here, $m$ is the mass and $v$ is the velocity.

The potential energy is given by,

$U = mgh$

Here, $m$ is the mass, $g$ is the acceleration due to gravity and $h$ is the height.

Since the girl starts from the bottom the initial potential energy will be zero and the final potential energy will be $mgh$.

Substitute $\frac{1}{2}m{v_{\rm{i}}}^2$for ${K_{\rm{i}}}$, $0{\rm{ J}}$ for ${U_{\rm{i}}}$, $mgh$ for ${U_{\rm{f}}}$and $\frac{1}{2}m{v_f}^2$for ${K_{\rm{f}}}$ in the above equation.

$\begin{array}{c}\\ - mgh = \frac{1}{2}m{v_{\rm{f}}}^2 - \frac{1}{2}m{v_{\rm{i}}}^2\\\\{v_{\rm{f}}} = \sqrt {{v_{\rm{i}}}^2 - 2gh} \\\end{array}$

Substitute ${\rm{8}}{\rm{.00 m / s}}$ for ${v_{\rm{i}}}$, $2.00{\rm{ m}}$for $h$ and $9.8{\rm{ m / }}{{\rm{s}}^2}$ for $g$in the above equation.

$\begin{array}{c}\\{v_{\rm{f}}} = \sqrt {{{\left( {8.00{\rm{ m / s}}} \right)}^2} - 2\left( {9.8{\rm{ m / }}{{\rm{s}}^2}} \right)\left( {2.00{\rm{ m}}} \right)} \\\\ = 4.98{\rm{ m / s}}\\\end{array}$

(B)

Apply the conservation of momentum.

${m_{{\rm{girl}}}}{u_{{\rm{girl}}}} + {m_{{\rm{box}}}}{u_{{\rm{box}}}} = \left( {{m_{{\rm{girl}}}} + {m_{{\rm{box}}}}} \right)v$

Here, ${m_{{\rm{girl}}}}{\rm{ and }}{m_{{\rm{box}}}}$ is the mass of girl and the box respectively, ${u_{{\rm{box}}}}{\rm{ and }}{u_{{\rm{girl}}}}$ is the initial velocity of box and girl respectively and $v$ is the final velocity of box and girl together.

Substitute $60.0{\rm{ kg}}$ for ${m_{{\rm{girl}}}}$, $15.0{\rm{ kg}}$ for ${m_{{\rm{box}}}}$, $4.98{\rm{ m / s}}$ for ${u_{{\rm{girl}}}}$and $0{\rm{ m / s}}$ for ${u_{{\rm{box}}}}$ in the above equation.

$\begin{array}{c}\\\left( {60.0{\rm{ kg}}} \right)\left( {4.98{\rm{ m / s}}} \right) + \left( {15.0{\rm{ kg}}} \right)\left( {0{\rm{ m / s}}} \right) = \left( {60.0{\rm{ kg}} + 15.0{\rm{ kg}}} \right)v\\\\v = \frac{{\left( {60.0{\rm{ kg}}} \right)\left( {4.98{\rm{ m / s}}} \right) + \left( {15.0{\rm{ kg}}} \right)\left( {0{\rm{ m / s}}} \right)}}{{\left( {60.0{\rm{ kg}} + 15.0{\rm{ kg}}} \right)}}\\\\ = 3.98{\rm{ m / s}}\\\end{array}$

(C)

Apply the law of conservation of energy, which says the gain in potential energy is equal to the loss in kinetic energy.

${U_{\rm{i}}} - {U_{\rm{f}}} = {K_{\rm{f}}} - {K_{\rm{i}}}$

Here, ${U_{\rm{i}}}{\rm{ and }}{U_{\rm{f}}}$ are the initial and final potential energy respectively and ${K_{\rm{i}}}{\rm{ and }}{K_{\rm{f}}}$ are the initial and final kinetic energy respectively.

The height will be calculated from the moment the girl grabs the box and reach to a height ${h_{{\rm{max}}}}$above the ground. Due to this, the final velocity of the girl at the maximum height will be $0{\rm{ m / s}}$.

Let the height where the girl grabs the box be $h$ and the maximum height reached by the girl be ${h_1}$.

Substitute $\frac{1}{2}m{v_{\rm{i}}}^2$for ${K_{\rm{i}}}$, $mgh$ for ${U_{\rm{i}}}$, $mg{h_1}$ for ${U_{\rm{f}}}$ and $\frac{1}{2}m{v_{\rm{f}}}^2$for ${K_{\rm{f}}}$ in the above equation.

$\begin{array}{c}\\mgh - mg{h_1} = \frac{1}{2}m{v_{\rm{f}}}^2 - \frac{1}{2}m{v_{\rm{i}}}^2\\\\h - {h_1} = \frac{{\left( {{v_{\rm{i}}}^2 - {v_{\rm{f}}}^2} \right)}}{{2g}}\\\end{array}$

Substitute ${\rm{8}}{\rm{.00 m / s}}$ for ${v_{\rm{i}}}$, $0{\rm{ m / s}}$for ${v_{\rm{f}}}$ and $9.8{\rm{ m / }}{{\rm{s}}^2}$ for $g$in the above equation.

$\begin{array}{c}\\h - {h_1} = \frac{{\left( {{{\left( {{\rm{3}}{\rm{.98 m / s}}} \right)}^2} - {{\left( {0{\rm{ m / s}}} \right)}^2}} \right)}}{{2\left( {9.8{\rm{ m / }}{{\rm{s}}^2}} \right)}}\\\\ = 0.808{\rm{ m}}\\\end{array}$

The total height reached by the girl will be,

$\begin{array}{c}\\{h_{{\rm{max}}}} = \left( {h - {h_1}} \right) + h\\\\ = \left( {0.808{\rm{ m}}} \right) + \left( {2.00{\rm{ m}}} \right)\\\\ = 2.81{\rm{ m}}\\\end{array}$

Ans: Part A

The speed immediately before the girl grabs the box is $4.98{\rm{ m / s}}$.

Part B

The speed immediately after the girl grabs the box is $3.98{\rm{ m / s}}$.

Part C

The maximum height the girl reaches is $2.81{\rm{ m}}$.

Answer 2

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