Question

A toy cork gun contains a spring whose spring constant is 10.0N/m. The spring is compressed 5.00cm and then used to propel a 6.00-g cork. The cork, however, sticks to the spring for 1.00cm beyond its unstretched length before separation occurs. The muzzle velocity of this cork is:A. 1.02m/s

B. 1.41m/s
C. 2.00m/s
D. 2.04m/s
E. 4.00m/s

Answer 1

2 m/s

Step-by-step explanation:

Spring constant, k = 10 N/m

Mass of the cork, m = 6 g = 0.006 kg

Initial position of the spring, x = 5 cm = 0.05 m

Final position of the spring, x' = 1 cm = 0.01 m

Let v be the muzzle speed of the cork.

According to the law of conservation of energy,

Initial potential energy = final potential energy + kinetic energy

0.5 x 10 x 0.05 x 0.05 = 0.5 x 10 x 0.01 x 0.01 + 0.5 x 0.006 x v²

0.0125 = 0.0005 + 0.003v²

v = 2 m/s

Thus, the muzzle speed of the cork is 2 m/s.

Answer 2

The muzzle velocity of this cork is 2 m/s.

Step-by-step explanation:

It is given that,

Spring constant of the spring, k = 10 N/m

Mass of the cork, m = 6 g = 0.006 kg

Initial position of the spring, x = 5 cm = 0.05 m

Final position of the spring, x' = 1 cm = 0.01 m

According to the law of conservation of energy, the initial potential energy of the spring is equal to the sum of final spring potential energy and the kinetic energy of cork such that,

`(1)/(2)kx^2=(1)/(2)kx'^2+(1)/(2)mv^2`

v is the muzzle velocity of this cork.

`kx^2=kx'^2+mv^2`

`v=\sqrt{(k(x^2-x'^2))/(m)}`

`v=\sqrt{(10* ((0.05)^2-(0.01)^2))/(0.006)}`

v = 2 m/s

So, the muzzle velocity of this cork is 2 m/s. Hence, this is the required solution.