impulse momentum and energy

Question

A 150-g block on a horizontal frictionless table is firmly attached to one end of a spring (of negligible mass) with the spring constant k=30 N/m. The other end of the spring is anchored to the wall. A 50-g ball is thrown horizontally toward the block with a speed of 5.0 m/s. The collision between the block and the ball is perfectly elastic. Find: a) the speed of the ball immediately after the collision; b) the maximum compression of the spring

Answer 1

perfectly elastic collision means the momentum and the total mechanical energy is conserved.

Let m = mass of the ball = 50 g = 0.05 kg

Let v₀ = initial velocity of the ball = 5 m/s

Initial momentum = mv₀ = 0.05*5 = 0.25 kg*m/s

Let v = final velocity of the ball

Let V = velocity of the block right after collision.

Let M = mass of the block = 150 g = 0.15 kg

 

Then conservation of momentum says:

MV - mv = 0.25 

and conservation of energy says:

0.5*MV² + 0.5*mv² = 0.5*mv₀² = 0.5*0.05*5*5 = 0.625 J

 

Now we have 2 equations and 2 unknowns which can be used to solve for V and v

 

Once you find the answer for part a) then you can use the conservation of energy equation to solve for maximum compression, A

0.5MV² = 0.5kA²

where k = 30N/m