Question

A group of students conduct the ballistic pendulum experiment, the same one we will do. The mass of their projectile was 70 grams, the mass of the block/pendulum was 250 grams. The true/effective length of the pendulum was 35 cm. The experiment was done on Earth where the local acceleration due to gravity is 9.80 m/s2. The pendulum swung up to a 38° angle. What was the speed of the projectile out of the cannon before it collides with the pendulum? Answer in units of meters per second, and to the fourth decimal place.

Answer 1

Initial energy of the system = Kinetic energy of (projectile+pendulum)
= (1/2)(m+M)v²
where m is the mass of projectile = 70 grams
M is the mass of pendulum = 250 grams
v is the initial velocity after the collision of both mass
Now when they swung up at the 38 degree then
Final energy of the system
= (m+M)gh
where h is height from the mean position = L - LCos = L(1-Cos) = 0.35(1-Cos38) = 0.0742 m
Now applying energy conservation
(1/2)(m+M)v² = (m+M)gh
v² = 2gh
v² = 2*9.81*0.0742
v = 1.206 m/s
Now
Consider momentum before the collision = mu + M*0 = mu
where u is the velocity of the projectile before collision
Final momentum of the system = (m+M)v
Applying conservation of momentum
(m+M)v = mu
u = (70+250)*1.206 /70 = 5.513 m/s