Physics problem help --- pendulum problem involving a collision, angle, mass and speed?

Question

A pendulum consists of a 1.00 kg bob attached to a string of length 2.10 m. A block of mass m rests on a horizontal frictionless surface. The pendulum is released from rest at an angle of 53° with the vertical and the bob collides elastically with the block. Following the collision, the maximum angle of the pendulum with the vertical is 5.73°. Determine the two possible values of the mass m.

Picture to go along with problem: http://i284.photobucket.com/albums/ll28/bathtub2008/8_65.gif

______kg (smaller value)

______kg (larger value)

Answer 1

Let's solve this problem in the general case, letting d be the length of the pendulum, b being the mass of the bob, m being the unknown mass, i being the starting angle and f being the final angle.  If we take the bottom of the swing as the point of zero potential energy for the bob, the system starts with a potential energy of bgd(1 − cos i), which represents the total energy in the system.  Because the collision is perfectly elastic, this represents the total energy after collision as well.  After collision, the bob still has an energy of bgd (1 − cos f).  At the bottom of the pendulum's swing just before collision, the entire energy of the system is in kinetic energy.

If pi is the momentum of the bob just before collision, pf is the momentum of the bob just after collision, and pm is the momentum of the mass being struck, then

pi = pm + pf, but pf can be either negative or positive.  pm is therefore pi ± |pf|.  If you calculate the kinetic energy in terms of momentum rather than speed, you get

K = ½p²/M, where M is the mass of the object for which the calculation is being done.  Besides getting |p| = √(2KM), we also get the second equation

½pi²/b = ½pf²/b + ½pm²/m

= ½pf²/b + ½(pi ± |pf|)²/m

= ½pf²/b + ½pi²/m  Ã‚± pi|pf|/m + ½pf²/m

or, in terms of energies Ei = ½pi²/b and Ef = ½pf²/b, this becomes

Ei = Ef + Ei*b/m + Ef*b/m ± 2*b/m√(EiEf)

Rearranging things gives

m = b(Ei + ± 2*√(EiEf) + Ef) / (Ei − Ef)

or, if you prefer,

= b(√Ei ± √Ef)² / ((√Ei)² − (√Ef)²)

= b(√Ei + √Ef) / (√Ei − √Ef) or b(√Ei − √Ef) / (√Ei + √Ef)

But

Ei = bgd(1 − cos i) and

Ef = bgd(1 − cos f)

The factor bgd cancels from the equation for m, meaning that the acceleration of gravity g and the length of the string d are entirely irrelevant.

If we introduce variables hi and hf such that hi² = 1 − cos i and hf² = 1 − cos f, the formula reduces to

m = b (hi + hf) / (hi − hf) or

m = b (hi − hf) / (hi + hf)

For this particular problem,

hi² = 1 − cos 53°= 0.398 and

hf² = 1 − cos 5.73° = 0.00500

so hi = 0.631, hf = 0.071, hi + hf = 0.702, hi − hf = 0.560, and

m = 1.00 (0.702/0.560) = 1.25 kg or

m = 1.00 (0.560/0.702) = 0.798 kg