Question

1.A block of mass 2.0 kg is pushed down on a vertically oriented spring that has a spring constant of 1500N/m.. Initially, the spring is compressed by 5.0 cm and the block is at rest. When the block is released, it accelerates upward. Find the speed of the block when the spring has returned to its equilibrium position.

2.The collision between a hammer and a nail can be considered to be approximately elastic. Assume that the nail, initially at rest, is struck by a 500-g hammer moving with an initial speed of 5.0 m/s. If the speed of the hammer is 4.6 m/s after the collision, calculate the following:

a.Kinetic energy of the nail after the collision

b.Momentum of the nail after the collision

c.The mass of the nail.

Answer 1

Here, m = 2.0 kg.

k = 1500 N/m

x = 5 cm = 0.05 m

Suppose the speed of the block is v m/s when the spring has returned to its equilibrium condition.

Now, apply conservation of energy -

(1/2)mv^2 = (1/2)kx^2

=> v = x*sqrt(k/m) = 0.05*sqrt(1500/2) = 1.37 m/s.

(2) Here, first apply conservation of momentum -

m1u1 + m2u2 = m1v1 + m2v2

=> 0.5*5 + m2*0 = 0.5*4.6 + m2*v2

=> m2v2 = 2.5 - 2.3 = 0.20

=> m2 = 0.2/v2---------------------(i)

collision is elastic, so kinetic energy shall be conserved.

so, (1/2)m1u1^2 + (1/2)m2u2^2 = (1/2)m1v1^2 + (1/2)m2v2^2

=> 0.5*5^2 + 0 = 0.5^4.6^2 + m2v2^2

=> m2v2^2 = 1.18---------------------------------------------------(ii)

from (i) and (ii) -

v2 = 1.18/0.2 = 5.9 m/s

and m2 = 0.2/5.9 = 0.034 kg.

(a) kinetic energy of the nail after collision = (1/2)m2v2^2 = 0.5*0.034*5.9^2 = 0.592 J

(b) Momentum of the nail = m2v2 = 0.034*5.9 = 0.20 kg-m/s

(c) mass of the nail = 0.034 kg