Question

A 0.50-kilogram cart is rolling at a speed of
0.40 meter per second. If the speed of the cart is
doubled, the inertia of the cart is
(1) halved (3) quadrupled
(2) doubled (4) unchanged

Answer 1

     The inertia of moving body is calculated from the amount of moviment equation, given by:

$Q=mv$
 
     Let us consider the index 1 to the initial situation, and the index 2 to the final situation. Doing this, we get:

 $\left \{ {{Q_{1}=mv_{1}-\ \textgreater \ Q_{1}=0.5*0.4-\ \textgreater \ Q_{1}=0.2 \frac{kg*m}{s} } \atop {Q_{2}=mv_{2}-\ \textgreater \ Q_{2}=0.5*2*0.4-\ \textgreater \ Q_{1}=0.4 \frac{kg*m}{s}}} \right.$
  
     Comparing the results:

$k= \frac{Q_{2}}{Q_{1}} \\ k= \frac{0.4}{0.2} \\ \boxed {k=2}$
 
     Therefore, the inertia is doubled.

Number 2

If you notice any mistake in my english, please let me know, because i am not native.

Answer 2

If the speed of the cart is  doubled, the inertia of the cart is doubled as well. Option 2 is correct.

The inertia of a body is the product of the mass of a body and its velocity. Mathematically I = mv

If a 0.50-kilogram cart is rolling at a speed of  0.40 meter per second, the inertia is expressed as;

$I_1 = 0.50\times 0.40$

$I_1=0.2kgm/s$

If the speed is doubled, the new speed will be 0.80m/s. The new inertia will be expressed as:

$I_2=0.8 \times 0.5\\I_2= 0.4kgm/s$

Taking the ratio of the inertia:

$\frac{I_2}{I_1}=\frac{0.4}{0.2}\\ \frac{I_2}{I_1}=2\\I_2=2I_1$

Hence if the speed of the cart is  doubled, the inertia of the cart is doubled as well.

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