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
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
To find the initial velocity of the gum, we can use the principle of conservation of mechanical energy. The initial kinetic energy of the gum will be equal to the potential energy gained by the pendulum mass.
First, let's find the potential energy gained by the pendulum mass. The formula for potential energy is given by:
PE = mgh
Here, m is the mass of the pendulum (1.62 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical height gained by the pendulum mass (12.7 cm or 0.127 m).
PE = m * g * h
= 1.62 kg * 9.8 m/s^2 * 0.127 m
≈ 2.17 J (rounded to two decimal places)
Since the gum sticks to the pendulum mass, the potential energy gained by the pendulum is equal to the initial kinetic energy of the gum.
The formula for kinetic energy is given by:
KE = (1/2) * m * v^2
Here, m is the mass of the gum (9 grams or 0.009 kg), and v is the initial velocity of the gum.
Setting the potential energy gained equal to the kinetic energy of the gum, we can solve for v:
2.17 J = (1/2) * 0.009 kg * v^2
v^2 = (2.17 J) / (0.009 kg * (1/2))
v^2 ≈ 480.89 m^2/s^2 (rounded to three decimal places)
Taking the square root of both sides, we find:
v ≈ 21.94 m/s (rounded to two decimal places)
Therefore, the initial velocity of the wad of gum is approximately 21.94 m/s.