Posted by **Jacob** on Monday, April 12, 2010 at 11:55pm.

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

- physics -
**drwls**, Tuesday, April 13, 2010 at 8:17am
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

286 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

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