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In the figure below, two particles are launched from the origin of the coordinat

ID: 2004142 • Letter: I

Question

In the figure below, two particles are launched from the origin of the coordinate system at time t = 0. Particle 1 of mass m1 = 5.00 g is shot directly along an x axis (on a frictionless floor), where it moves with a constant speed of 14.0 m/s. Particle 2 of mass m2 = 3.80 g is shot with a velocity of magnitude 28.0 m/s, at an upward angle such that it always stays directly above particle 1 during its flight.
(a) What is the maximum height Hmax reached by the center of mass of the two-particle system?
____m
(b) What is the magnitude and direction of the velocity of the center of mass when the center of mass reaches this height?
___m/s
___° (counterclockwise from the positive x axis)

(c) What is the magnitude and direction of the acceleration of the center of mass when the center of mass reaches this height?
__ m/s^2
___° (counterclockwise from the positive x axis)

Explanation / Answer

The particle 2 is always stays above the paricle one means the shadow's of these two are coinside
v_2x = v_1 = 14 m/s iniital velocity v _2 = 28 m/s v_2 = sqrt ( v_2x)^2 + (V_2y)^2 verticle component of velocity v _2y = sqrt ( (v_2)^2 - (v_2x)^2 )    = sqrt(588) m/s Hence the maximum height y_max for paricle 2    = (v_2y)^2 /2g   = 588 / 19.6 = 30 m the maximum height reached by the center of mass H_max =   m_2 *y_max / (m_1 + m_2 )    .............(1) here mass of paricle 1 m_1 = 5 g and    m_2 = 3.8 g plug all valuee (1) we get H_max =12.95 m ----------------------------------------------- (b) since the both particles have the same horizontal velocity , at this maximum height the vetricle component of particle 2 is zero .hence the velocity of the centre of mass is 14 m/s along positive X axis direction 0^0 ----------------------- (c) the particle 2 willl experience the acceleration due to gravity   g = 9.8 m/s hence the the accelaration of the centre of mass at this height a   = m_2 * g / (m_1 + m_2)   =   4.23 m/s^2 The direction is vertically down direction 270^0   countre clock wise from the positive x-axis