Answer:
1. [tex]t_2 = 2t_1[/tex]
2. [tex]t_2 = t_1\sqrt{2}[/tex]
Explanation:
1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.
In Newton's 2nd law: F = ma
where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.
[tex]a_1 = 2a_2[/tex]
Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time
[tex]t = v / a_1[/tex]
Now that acceleration is halved:
[tex] t = \frac{v}{2a_2}[/tex]
[tex] \frac{v}{a_2} = 2t[/tex]
You would need to push for twice amount of time [tex]t_2 = 2t_1[/tex]
2. The distance traveled by the puck is as the following equation:
[tex]d = at^2[/tex]
So if the acceleration is halved while maintaining the same d:
[tex]\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}[/tex]
As [tex]d_1 = d_2[/tex], then [tex]d_1/d_2 = 1[/tex]. Also [tex]a_1 = 2a_2[/tex]
[tex]1 = \frac{2a_2t_1^2}{a_2t_2^2}[/tex]
[tex] t_2^2 = 2t_1^2[/tex]
[tex]t_2 = t_1\sqrt{2}\approx 1.14t_1[/tex]
So t increased by 1.14
