Question 3 One 2.5 kilogram piece of mass moving at 100 meters per second (m/s) collides and sticks to another 2.5 kilogram mass at rest. The resulting 5 kilogram mass continues in the same direction with some speed. No external forces act on these masses, and during the collision no particles are absorbed, but 3 × 1025 photons are emitted due to the collision. These photons all have a wavelength of 1.0 nm, and all of them move in the opposite direction of the initial 2.5 kilogram mass moving at 100 m/s. Using momentum conservation, compute the speed of the resulting 5 kilogram mass. Answer in meters per second, and to the nearest fourth decimal place.
Question 5 One 2.5 kilogram piece of mass moving at 100 meters per second (m/s) collides and sticks to another 2.5 kilogram mass at rest. The resulting 5 kilogram mass continues in the same direction with some speed. No external forces act on these masses, and during the collision no particles are absorbed, but 3 × 1025 photons are emitted due to the collision. These photons all have a wavelength of 1.0 nm, but their direction is split three ways. One third move in a direction opposite the 2.5 kilogram mass moving at 100 m/s. The remaining photons are emitted perpendicular to the direction of motion of the resulting mass. Using momentum conservation, compute the speed of the resulting 5 kilogram mass. Answer in meters per second, and to the nearest fourth decimal place.
Question 3 :
m1 = mass of the moving piece of mass before collision = 2.5 kg
v1i = initial velocity of the mass moving before collision = 100 m/s
m2 = mass of the piece of mass at rest before collision = 2.5 kg
v2i = initial velocity of the mass at rest before collision = 0 m/s
v = velocity of the two masses together after collision
n = number of photons = 3 x 1025
= wavelength of each photon = 1 x 10-9 m
Using conservation of momentum
m1 v1i + m2 v2i = (m1 + m2) v - nh/
(2.5 x 100) + (2.5 x 0)= (2.5 + 2.5) v - (3 x 1025 ) (6.63 x 10-34)/(1 x 10-9)
v = 53.978 m/s
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