Question

Problem 1 An object with m1 = 5kg is attached to a spring of negligible mass. This mass/spring combination is then slid horizontally on a frictionless surface with a velocity of 5 m/s towards a stationary object with m2 = 6kg. Upon impact, the spring compresses, then we examine two cases. First, find the velocities of the two objects assuming the spring completely relaxes again after the interaction. Second, assume that m2, after they separate, slides up a frictionless incline. (a)What is the relative speed of the masses when the spring is maximally compressed?

Answer 1

Answer and Explanation:

The answer is attached below

Answer 2

The velocity of both the objects after the collision is $\frac{25}{11} m/s$.

(a) the relative velocity of the masses is zero.

(b) the compression of the spring is 0.185m.

Inelastic collision:

The given case is an example of an inelastic collision in which two objects after the collision move together. The momentum of the system is conserved, therefore,

$m_1u_1+m_2u_2=(m_1+m_2)v$

here, m₁ = 5kg , m₂ = 6kg

u₁ = 5 m/s , initial velocity of mass m₁, and

u₂= 0, initial speed of the mass m₂

$5\times5+0=(5+6)v\\\\v=\frac{25}{11}m/s$

(a) When the spring is fully compressed both the masses move with the same velocity, therefore the relative speed of the masses is zero.

(b) from the law of conservation of energy:

the initial kinetic energy of the masses is converted into final kinetic energies and the potential energy of the spring:

$\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u^2_2=\frac{1}{2} (m_1+m_2)v^2+\frac{1}{2}kx^2$

where x is the compression of the spring

$\frac{1}{2}\times5\times5^2+0=\frac{1}{2}(5+6)\times(\frac{25}{11})^2=\frac{1}{2}\times2000x^2\\\\x=0.185m$

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