1) A 14 gram clay mass moving at 50 m/s collides into an 86 gram clay mass at re
ID: 1284444 • Letter: 1
Question
1) A 14 gram clay mass moving at 50 m/s collides into an 86 gram clay mass at rest. There are no external forces acting on these masses during the collision. What is the speed in m/s of the resulting mass post collision?
5) A group of students conduct the ballistic pendulum experiment, the same one we did. 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 were the local acceleration due to gravity is 9.8 m/s^2. The pendulum swung up an angle of 38 degrees.
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.
1) A 14 gram clay mass moving at 50 m/s collides into an 86 gram clay mass at rest. There are no external forces acting on these masses during the collision. What is the speed in m/s of the resulting mass post collision? 2)What is the ratio of the post-collision kinetic energy of the system to the pre-collision kinetic energy of the system, 3) Issac Newton refered to momentum as the quantity of motion, and in classical physics it is defined as P = mV Where p is the momentum vector mm is the mass, and v is the velocity vector. What are the standard SI units of 4) The kinetic energy of a particle can be written Where m is its mass and v is its speed. In terms of mass and speed, what are the dimensions of kinetic energy? mass^____ * speed^____ (Answers choices per blank are: 1,0,-1,2,-2) 5) A group of students conduct the ballistic pendulum experiment, the same one we did. 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 were the local acceleration due to gravity is 9.8 m/s^2. The pendulum swung up an angle of 38 degrees. 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.Explanation / Answer
1) 14*50 = (14 + 86)*v
v=7 m/s
2) Kpost/Kpe = 0.5*100*7^2/(0.5*14*50^2)= 0.14
3) units are kg^1 m^1 s^-1
4)
mass^1 speed^2
5)
Ei = Ef
1/2 m v^2 = m g h = m g L( 1- cos theta)
0.5*v^2 = 9.81*.35*(1-cos(38 degrees))
v=1.207 m/s
now conservation of momentum
70*v = 250*1.207
v=4.3089 m/s
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.