A block of mass m1 travels at a speed of v0 on a frictionless horizontal surface

Question

A block of mass m1 travels at a speed of v0 on a frictionless horizontal surface when it comes upon a second block of mass m2 which is initially motionless. Block m2 has a massless spring with spring constant k in front of it. a. Explain why the linear momentum of the system of two blocks and spring is or is not conserved during the collision. b. Explain why the mechanical energy of the system of two blocks and spring is or is not conserved during the collision. c. Consider a time when the spring is compressed a distance x from its equilibrium length. Write the equations for the conservation of momentum and conservation of energy at this instant. d. Now consider the instant when the spring is compressed as far as it will ever be compressed, a displacement L from its equilibrium length. What are the velocities of the two masses at this instant? e. What is the spring’s maximum compression L in terms of m1, m2, k, and v0, or a subset of these quantities? f. In the end, when block m1 has been pushed away and is no longer touching the spring, what is the final velocity of block m1? I understand and have gotten the answers for a-d, but don't know how to solve for e and f.

Explanation / Answer

a. Linear momentum is conserved because no external forces (in the direction of motion) are applied.

b. Assuming the spring is "ideal," there is no (permanent) deformation and no losses to friction, so energy is conserved.

c. At any point in time, the initial kinetic energy is equal to the spring energy and the current kinetic energy, or

½*m1*v0² = ½kx² + ½*m1*v1² + ½*m2*v2²

and of course you could cancel the "½" terms if you like.

d. At maximum compression, the two masses have the same velocity (since the spring is neither compressing nor stretching at that point), so

½*m1*v0² = ½kL² + ½(m1+m2)*v1²

Conservation of momentum tells us that

m1*v0 = (m1+m2)*v1, or

v0 = (m1+m2)*v1/m1

Therefore

½*m1*((m1+m2)*v1)/m1)² = ½kL² + ½(m1+m2)*v1² multiply by 2

m1*(((m1+m2)*v1)/m1)² = kL² + (m1+m2)*v1²

(m1+m2)²*v1²/m1 = kL² + (m1+m2)*v1²

v1²*((m1+m2)²/m1 - (m1+m2)) = kL²

v1²*(m1+m2)((m1+m2)/m1 - 1) = kL²

v1²*(m1+m2)((1 + m2/m1 - 1) = kL²

v1²*(m1+m2)(m2/m1) = kL²

v1² = kL²*m1 / (m2*(m1+m2))

v1 = L*(k*m1 / (m2*(m1+m2)))

e. Starting from this point: v1²*(m1+m2)(m2/m1) = kL²

and substituting with v1 = v0*m1/(m1+m2)

gives us

v0²*m1²*(m1+m2)*m2 / (m1*(m1+m2)²) = kL²

v0²*m1*m2 / (m1+m2) = kL²

L² = v0²*m1*m2 / k*(m1+m2)

L = v0*(m1*m2 / k*(m1+m2))

f. Once x = 0 again, ½*m1*v0² = 0 + ½*m1*v1² + ½*m2*v2²

Multiplying by 2 and substituting

v2 = v0 + v1 (from conservation of energy and momentum) gives

m1*v0² = m1*v1² + m2*(v0+v1)² = m1*v1² + m2*v0² + m2*2*v0*v1 + m2*v1²

0 = (m1+m2)*v1² + 2*m2*v0*v1 + (m2 - m1)*v0²

This is quadratic in v1 and so

v1 = (-2*m2*v0* ± [4*m2²*v0² - 4*(m1+m2)(m2-m1)*v0²]) / 2(m1+m2)

v1 = (-2*m2*v0* ± 2*v0*[m2² + (m1+m2)(m1-m2)]) / 2(m1+m2)

v1 = (-m2*v0* ± v0*[m2² + (m1² - m2²)]) / (m1+m2)

v1 = (-m2*v0* ± v0*m1) / (m1+m2) = v0*(-m2 ± m1) / (m1+m2)

We can further reason that the "-" in "±" can be disregarded, since that would yield v1 = -v0, which is possible ONLY if m2 has infinite mass (and v2 = 0)

v1 = v0*(-m2 + m1) / (m1 + m2) = v0*(m1 - m2) / (m1 + m2)

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