Question

(worth two problems!) A massless spring of spring constant 3800N/m is compressed between two blocks (it

Answer 1

(a)The original potential energy of the system is equal to energy stored in the spring.

If x is compression distance of the spring

PE =1/2kx² =1/2 (3800 N/m) x²=1900x2

(b)The initial momentum of the system is,

= (1.40 kg) (2.90 m/s)+(2.0 Kg)(2.90 m/s)

= 9.86 kgm/s

(c)Using the conservation of linear momentum along the x axis,

9.86 kg . m/s = (1.40 kg)(ツcos θ)-(2.0 kg)(3.5 m/s)cos 34° w cos θ)-2.897 m/s 0+0-m, ("Y sin θ)-m, ( ar sin 34°) Solve the eequation for cos θ Use the conservation of linear momentum al ong the y azis, then sin 34° (2.0 kg) (3.5 m/s) sin 34° 140 kg 2.795 m/s THerefore y,-by sin θ)' +(nf cose).-M2.795 m/s)" + (2.897 m/s)' - 4.02 m/s And the directi on is, 2.795 m 4.02 m/s =46.02° below horizontal -1 e-tan