Question

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 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:

  Second  projectile is 1.4 times faster than first projectile.

Explanation:

By linear momentum conservation

Pi = Pf

m x U + M x 0 = (m + M) x V

$U= \dfrac{(m + M)\times V}{m}$

Now Since this projectile + pendulum system rises to height 'h', So using energy conservation:

KEi + PEi = KEf + PEf

PEi = 0, at reference point

KEf = 0, Speed of system zero at height 'h'

$KEi = \dfrac{(m + M)\times V^2}{2}$

PEf = (m + M) g h

So,

$\dfrac{(m + M)\times V^2}{2} + 0 = 0+ (m + M) g h$

$V =\sqrt {2gh}$

So from above value of V

Initial velocity of projectile =U

$U=\dfrac{(M+m)\sqrt{2gh}}{m}$

Now Since mass of projectile and pendulum are constant, So Initial velocity of projectile is proportional to the square root of height swung by pendulum.

Which means

$\dfrac{U_2}{U_1}=\sqrt{\dfrac{h_2}{h_1}}$

$U_2=\sqrt{\dfrac{h_2}{h_1}}\times U_1$

$U_2=\sqrt{\dfrac{5.2}{2.6}}\times U_1$

U₂ = 1.41 U₁

Therefore we can say that ,Second  projectile is 1.4 times faster than first projectile.