Mass on a spring: A 2.00-kg object is attached to an ideal massless horizontal spring of spring constant 100.0 N/m and is at rest on a frictionless horizontal table. The spring is aligned along the x-axis and is fixed to a peg in the table. Suddenly this mass is struck by another 2.00-kg object traveling along the x-axis at 3.00 m/s, and the two masses stick together. What are the amplitude and period of the oscillations that result from this collision?
A) 0.300 m, 1.26 s
B) 0.300 m, 0.889 s
C) 0.424 m, 0.889 s
D) 0.424 m, 1.26 s
E) 0.424 m, 5.00 s

Question

Quizplus · Accepted Answer

Explanation:The momentum of the system is conserved, so the total momentum before the collision is equal to the total momentum after the collision. The initial momentum is:$$p_i = m_1v_1 = (2.00	ext{ kg})(3.00	ext{ m/s}) = 6.00	ext{ kg m/s}$$The final momentum is:$$p_f = (m_1 + m_2)v_f = (2.00	ext{ kg} + 2.00	ext{ kg})v_f = 4.00	ext{ kg}v_f$$Setting these two momenta equal to each other, we get:$$6.00	ext{ kg m/s} = 4.00	ext{ kg}v_f$$Solving for $v_f$, we get:$$v_f = 1.50	ext{ m/s}$$The amplitude of the oscillations is equal to the maximum displacement of the mass from its equilibrium position. The maximum displacement occurs when the mass is at the end of its swing, and its velocity is zero. The maximum displacement is:$$A = \frac{v_f}{\omega} = \frac{1.50	ext{ m/s}}{\sqrt{100.0	ext{ N/m}/4.00	ext{ kg}}} = 0.300	ext{ m}$$The period of the oscillations is equal to the time it takes for the mass to complete one full cycle of motion. The period is:$$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{4.00	ext{ kg}}{100.0	ext{ N/m}}} = 1.26	ext{ s}$$

Mass on a Spring: a 2