Question

A 3.0-kg block slides along a frictionless tabletop at 8.0 m/s toward a second block (at rest) of mass 4.5 kg. A coil spring, which obeys Hooke’s law and has spring constant k = 850 N/m, is attached to the second block in such a way that it will be compressed when struck by the moving block, Fig. 9-40. (a) What will be the maximum compression of the spring? (b) What will be the final velocities of the blocks after the collision? (c) Is the collision elastic? Ignore the mass of the spring.

Answer 1

This question involves the concepts of the law of conservation of energy and the law of conservation of momentum.

(a) The maximum compression of the spring will be "0.475 m".

(b) The final velocities of the blocks after collision will be "".

(c) The collision is "elastic".

(a)

The maximum compression can be obtained by converting all the kinetic energy of the block to the potential energy of the spring. According to the law of conservation of energy:

$\frac{1}{2}mv^2=\frac{1}{2}kx^2\\\\x=v\sqrt{\frac{m}{k}}$

where,

x = maximum compression = ?

v = speed of block = 8 m/s

m = mass of block = 3 kg

k = spring constant = 850 N/m

Therefore,

$x=(8\ m/s)\sqrt{\frac{3\ kg}{850\ N/m}}$

x = 0.475 m

(b)

Using the law of conservation of momentum:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = mass of block 1 = 3 kg

m₂ = mass of block 2 = 4.5 kg

u₁ = initial speed of block 1 = 8 m/s

u₂ = initial speed of block 2 = 0 m/s

v₁ = v₂ = final speeds of block 1 and 2 = v = ?

Therefore,

$(3\ kg)(8\ m/s)+(4.5\ kg)(0\ m/s)=(3\ kg+4.5\ kg)v\\\\v=\frac{24\ kg.m/s}{7.5\ kg}$

v = 3.2 m/s

(c)

Since friction is absent and the system follows both the law of conservation of energy and the law of conservation of momentum. Therefore, the collision is elastic.

Learn more about the law of conservation of energy here:

brainly.com/question/20971995?referrer=searchResults

The attached picture explains the law of conservation of energy.