Conservation of momentum and conservation of energy were applied to the same tim

Question

Conservation of momentum and conservation of energy were applied to the same time period in the previous problem. The time period was defined as the moment just before the collision to moment just after the collision. Momentum just before was an elastic collision. and energy were both conserved for this time period because it was an elastic collision. You may encounter other problems in which you will need to apply conservation of momentum to one part of the problem conservation of energy to another part These can be thought of as 'multi-stage' problems because they involve different stages, or time periods to consider. For example: A bullet collides with a stationary block, becomes embedded in the block, which then slides off a table and falls to the floor. Momentum is conserved during the collision. Energy is not conserved during the collision, but energy is conserved during the fall to the floor. A pendulum is pulled back to a certain angle and released. When it gets to the bottom of its swing, it collides with a soccer ball. Conservation of energy can be applied to the swinging pendulum, while conservation of momentum can be applied to the collision. Energy is conserved during the pendulum's swing, but it is unclear if energy is conserved during the collision, as no indication was given as to whether the collision was elastic. Use conservation of energy and momentum principles to solve the following problems A 5 g bullet travelling at 300 m/s strikes and is embedded in a 500 g stationary target. The bullet target move across a level frictionless table before striking a spring, which has a spring constant, k 1000 N/m. What is the maximum distance the spring can be compressed?

Explanation / Answer

m = mass of bullet = 5 g = 0.005 kg

v = speed of bullet before collision = 300 m/s

M = mass of target = 500 g = 0.5 kg

Vi = initial speed of target before collision = 0 m/s

V = speed of target+bullet after collision

using conservation of momentum

mv + MVi = (m + M) V

0.005 x 300 + 0.5 x 0 = (0.005 + 0.5) V

V = 2.97 m/s

x = compression of spring

using conservation of energy

spring potential energy = kinetic energy of bullet-target combination

(0.5) k x2 = (0.5) (m + M) V2

(1000) x2 = (0.005 + 0.5) (300)2

x = 6.74 m

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