A student throws a 10 gram of gum is thrown across the physics lab from a distan

Question

A student throws a 10 gram of gum is thrown across the physics lab from a distance of 3.3 m. When the gum has a horizontal velocity of v0 it connects with a mass on the end of 1.46 kg simple pendulum which is initially at rest. The pendulum has a length of L of 0.77 m. The gum sticks to the pendulum mass and swings backwards to gain a vertical height of 12.2 cm.

1)

What is initial velocity of the wad of gum?

vmax =

2)

How long does it take the pendulum (and the gum) to come to rest for the first time?

t =

I've been working on this one for hours and I can't seem to figure it out... if you are able to figure it out please show the work so I can figure out how to do it.

Explanation / Answer

If the pendulum/gum reached a vertical height of 12.2 cm (0.122m), we can determine its kinetic energy just after collision since the kinetic energy just after collision is equal to the potential energy when it reaches its high point

so we have

1/2 (m+M)v^2 = (m+M) gh where m is the mass of the gum, M the mass of the pendulum, and v the speed of the system after collision, therefore we have

v=sqrt[2gh]=sqrt[2x9.8x0.122]=1.546m/s

we use conservation of momentum to find V, the initial velocity of the gum

we know that

mV = (m+M)v or
V=(m+M)v/m = (0.01+1.46)(1.546)/0.01 = 227.262m/s
which is an incredibly high speed.

b) without friction, the system should not come to rest; to calculate when the system comes to rest would require more information about any fricitonal forces

if you mean how long it takes for the pendulum to return to its equilibrium position, then we calculate the period of the pendulum and divide that by 4 (since returning to the initial vertical position is 1/4 of a period)

the period of a simple pendulum is

P= 2pi sqrt[L/g]

set L=0.77 and get: P=2 pi sqrt[0.77/9.8]=1.761 s

1/4 th of this is 0.44s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.