posted by Jacob on .
A student throws a 9 gram of gum is thrown across the physics lab from a distance of 2.5 m. When the gum has a horizontal velocity of vo it connects with a mass on the end of 1.62 kg simple pendulum which is initially at rest. The pendulum has a length of L of 0.74 m. The gum sticks to the pendulum mass and swings backwards to gain a vertical height of 12.7 cm.
(a) What is initial velocity of the wad of gum?
vmax = ?
i used the formula 2pi (l/g)squareroot then 2pi/T to get the w and timed it by Amp but i still didn't get it right
You don't need the period of the pendulum (or the pendulum length) to do this problem. The vertical height H that the pendulum rises tells you the velocity the pendulum (mass M) and gum (mass m) acquire right after the collision. Call that velocity V.
V = sqrt (2 g H)= 1.58 m/s
Now apply conservation of momentum to the impact-and-sticking process (before the pendulum has swung at all). Let vo be the velocity of the gum before impact.
m vo = (M+m)V = (M+m)sqrt(2gH)
Solve for vo.
vo = (1.629/0.009)*1.58 m/s
I don't think gum can be thrown that fast. It is nearly the speed of sound. However, the method seems correct to me. I believe that unrealistic numbers have been provided