Question

A 15-kg block at rest on a horizontal frictionless surface is attached to a very light ideal spring of force constant 450 N/m. The other end of the spring is attached to a fixed wall. A lump of putty travels horizontally with a speed of 8.0 m/s towards the block from the side directly opposite the spring. The putty strikes and sticks to the block. What is the maximum distance the spring is compressed after the impact

Answer 1

Answer:

0.266 m

Explanation:

Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,

P= mv where p is momentum, m is mass and v is the speed of an object. In this case

$m_pv_p=v_c(m_p+m_b)$ where sunscripts p and b represent putty and block respectively, c is common velocity.

Substituting the given values then

3*8=v(15+3)

V=24/18=1.33 m/s

The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy

$0.5(m_p+m_b)v_c^{2}=0.5kx^{2}$ where k is spring constant and x is the compression of spring. Substituting the given values then

$(3+15)*1.33^{2}=450*x^{2}\\x\approx 0.266 m$