Question

HW-65 problem concerns three collision experiments performed on a frictionless surface with Tiers A and B, with masses mA and mg, respectively. In all three experiments, the instantaneous seeds of gliders A and B are initially w, and 0, respectively. If not given enough information is given to answer any of the parts, state so explicitly. a. In experiment 1, the track is level, and the gliders are repelled by magnets, so they do not come into contact. At time . glider A moves to the left and glider B to the right, both with speed 0./2, as shown at right. i. Is Ipsd, the magnitude of the momentum of the system of both gliders at time is greater than, less than or equal to mo.? Explain. il. Is IPad, the magnitude of the momentum of elider B at time to greater than, less than, or equal to lps, the magnitude of the momentum of glider A at time ? Explain. Per b. In experiment 2, the track is inclined, as shown. At times. glider A has an instantaneous speed of O. Is Bed, the magnitude of the momentum of the system of the two gliders at time in experiment 2. greater than, less than, or equal to my? Explain. c. In experiment 3, the track is again level, and glider B's magnet is replaced by a spring-loaded plunger. (The mass of glider B is unchanged.) The plunger is initially compressed, and it is released when the gliders come into contact. At time ir, glider A moves to the left with speed greater than D.. Is lod, the magnitude of the momentum of the system of the two gliders at time t, in experiment 3 greater than, less than, or equal to m.v.? Explain. 5 = ? HUUHB Pearson Custom Publishing

Answer 1

1. For experiment 1 ,at time ti, glider B is at rest so its corresponding velocity is zero. Hrnce, the initial momentum of the system will be due to glider A only and its magnitude can be given as

|p|= (m_A v_0).......(1)

so momentum is equal to m_A v_0.

((ii) Now both the gliders are moving in opposite direction with v0/2 speed and we know that the momentum is a vector quantity so the magnitude of final momentum |pf| of the system is less than m_A v_0 and can be given as

|p_f|=(m_A-m_B) (v_0)/2

But the final magnitude of momentum of glider A and glider B are equal and is given by m_A v_0/2 =m_B v_0/2 provided their masses are same. If m_B is bigger (smaller) than final momentum of B is greater (lesser) than the final mometum of glider A

(b) Firstly, we need to find the final speed of glider B by applying conservation of linear momentum as follows

p_i=p_f

m_A v_0+0=0+m_B v_B

Solving for v_B

v_B=(m_A v_0)/(m_B)......(2)

Final momentum of the system is

|p_f|=m_B v_B=m_A v_0 (Using Eq.(2))

So it is equal to m_A v_0.

(c) Solving as above we need to find v_B as

v_B=(m_A v_A)/(m_B)......(3)

Final momentum of the system is

|p_f|=m_B v_B=m_A v_A (Using Eq.(3))

As v_A is greater than v_0 so momentum is greater than m_A v_0.