Three masses slide across a frictionless table. Each is approaching the center o

Question

Three masses slide across a frictionless table. Each is approaching the center of the table. The initial velocity of mass A and mass B are shown. MA =0.025kg,MB =30g,MC =0.055kg All three masses arrive at the center of the table at the same time and stick together. (A) If after the three masses are stuck together, they are moving at 0.6 m/s in the +x direction, what must have been the magnitude and direction of the velocity of mass C? (15 pts) (B) What was the change in the kinetic energy of the three mass system as a result of the collision? (5 pts) ( C) What was the work done by the non-conservative forces during the collision (5 pts)

Explanation / Answer

Using Momentum Conservation,
In X axis,
mc * Vc * cos() - ma * Va - mb * Vb * cos(55) = (ma + mb + mc) * 0.6
0.055 * Vc *cos() - 0.025 * 1.75 - 30 * 10^-3 * 1*cos(55) = (0.055 + 0.025 + 30 * 10^-3 ) * 0.6
Vc * cos() = 2.31 ------1

In Y axis,
mc * Vc * sin() = mb * Vb * sin(55)
0.055 * Vc * sin() = 30 * 10^-3 * 1* sin(55)
Vc * sin() = 0.447 ---------2

Solving 1 & 2,
tan() = 0.447/2.31
= 10.95o

Vc = 0.447/sin(10.95)
Vc = 2.35 m/s

(B)
Kinetic Energy Initial = 1/2 * ma * Va^2 + 1/2 * mb * Vb^2 + 1/2 mc * Vc^2
Kinetic Energy Initial = 1/2 * 0.025 *  1.75^2 + 1/2 * 30 * 10^-3 * 1^2 + 1/2 * 0.055 * 2.35^2
Kinetic Energy Initial = 0.20515 J

Kinetic Energy Final = 1/2 *  (ma + mb + mc) * V^2
Kinetic Energy Final = 1/2 *  (0.055 + 0.025 + 30 * 10^-3 ) * 0.6^2
Kinetic Energy Final = 0.0198 J

Change in KE = 0.18535 J

(C)
Work done = Change in Kinetic Energy = 0.18535 J

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.