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castortr0y [4]
3 years ago
6

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. T

he 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
Physics
1 answer:
den301095 [7]3 years ago
7 0

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

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The change made between glass and air would be determined by:

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