Question

A glider of mass m is free to slide along a horizontal air track. It is pushed against a launcher at one end of the track. Model the launcher as a light spring of force constant k compressed by a distance x. The glider is released from rest.

(a) Show that the glider attains a speed of $\mathit{v}\mathbf{=}\mathit{x}{\left(\frac{{\displaystyle k}}{{\displaystyle m}}\right)}^{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}$.

(b) Show that the magnitude of the impulse imparted to the glider is given by the expression$\mathit{I}\mathbf{=}\mathit{x}{\mathbf{\left(}\mathbf{k}\mathbf{m}\mathbf{\right)}}^{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}$.

(c) Is more work done on a cart with a large or a small mass?

Short answer

(a) It is proved that the glider attains a speed of $v=x\sqrt{\frac{k}{m}}$.

(b) It is proved that the impulse imparted to the glider is given by the expression$I=x\sqrt{km}$.

(c) The mass makes no difference to the work.

Step-by-step solution

Concept

The impulse imparted to a particle by a net force$\sum \overrightarrow{F}$is equal to the time integral of the force.

$\overrightarrow{I}=\underset{t_i}{\overset{t_f}{\int }}\overrightarrow{F}dt$

where

$\overrightarrow{I}=$Impulse delivered to object

$\overrightarrow{F}=$Average friction force

$\Delta t=$Time interval in seconds.

Isolated System (Energy): The total energy of an isolated system will always be conserved, so

$\Delta E_{system}=0$

which can be written as

$\Delta K+\Delta U+\Delta E_{\mathrm{int}}=0$

where

$\Delta K=$Change in kinetic energy

$\Delta U=$Change in potential energy

$\Delta E_{int}=$Change in internal energy.

If no nonconservative forces act within the isolated system, the mechanical energy of the system is conserved, so

$\Delta E_{mech}=0$

which can be written as

$\Delta K+\Delta U=0$

Step 2:Calculation of the speed of the glider

The mass of the glider is m.

Part (a):

The mechanical energy of the isolated spring-mass system is conserved, then we can write

$\begin{array}{c}\Delta K+\Delta U=0\\ \text{Initial\hspace{0.17em}total\hspace{0.17em}energy}=\text{Final\hspace{0.17em}total\hspace{0.17em}energy}\\ \text{Initial\hspace{0.17em}Kinetic\hspace{0.17em}energy}+\text{Initial\hspace{0.17em}potential\hspace{0.17em}energy}=\text{Final\hspace{0.17em}Kinetic\hspace{0.17em}energy}+\text{Final\hspace{0.17em}potential\hspace{0.17em}energy}\\ K_i+U_{si}=K_f+U_{sf}\end{array}$

We know that the initial kinetic and final potential energies are zero; in this case, thus the above expression to calculate the speed v can be written as

$\begin{array}{c}0+\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2+0\\ v=x\sqrt{\frac{k}{m}}\end{array}$

Thus, it is proved that the glider attains a speed of $v=x\sqrt{\frac{k}{m}}$.

Step 3:The magnitude of the impulse imparted to the glider

Part (b):

We can calculate the impulse imparted on the glider as

$\begin{array}{c}I=\Delta P\\ =|{\overrightarrow{P}}_f-{\overrightarrow{P}}_i|\\ =mv-0\end{array}$

Substitute the value of v in the above expression, we get

$\begin{array}{c}I=mx\sqrt{\frac{k}{m}}\\ I=x\sqrt{km}\end{array}$

Thus, it is proved that the impulse imparted to the glider is given by the expression.

$I=x\sqrt{km}$

Work done on the cart

Part (c):

For the glider, the work done can be calculated as

$\begin{array}{c}W=K_f-K_i\\ =\frac{1}{2}m{v}^2-0\\ =\frac{1}{2}m{\left(x\sqrt{\frac{k}{m}}\right)}^2\\ =\frac{1}{2}k{x}^2\end{array}$

The mass makes no difference to the work done as it doesn’t depend on the factor mass.

Thus, work done is not affected by the mass.