Question

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

Answer 1

assuming you mean "how fast at the bottom of the swing", since the speed of a pendulum is constantly changing. The key with pendulums is the transfer of energy from potential to kinetic and back again as the pendulum oscillates. If m1 reaches a height of 2.cm, then\[m_1*g*(2.6cm) = \frac{1}{2} m_1 v^2\] where 2.6cm is hte height of the pendulum at its maximum. so the speed at the bottom is equal to \[v = \sqrt(2*g*2.6cm)\] If a second pendulum reaches twice as high, substitute twice the height into the same equation. It won't be twice as fast, because you're taking a square root, but it will be higher than the first speed. It's also worth seeting that the mass of the pendulum itself doesn's actually matter, since it cancels when we solve for velocity