Question

(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second projectile (of the same mass) causes the pendulum to swing twice as high, h2 = 5.2 cm. The second projectile was how many times faster than the first?

Answer 1

Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:  

$v = \frac{m + M}{m} \cdot \sqrt{2gh}$

where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum.  

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:    

$\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}$           (1)

where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively.  

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}$  

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}$

$\frac{v_{2}}{v_{1}} = 1.41$  

Therefore, the second projectile was 1.41 times faster than the first.  

I hope it helps you!