Particle A of mass m 0.120 kg moving with speed v 2.80 m/s along the positive x-
ID: 1864365 • Letter: P
Question
Particle A of mass m 0.120 kg moving with speed v 2.80 m/s along the positive x-direction strikes particle B, initially at rest, of mass m 0.140 kg in a glancing collision. As a result of the collision, particle A is deflected off at an angle of 30.0° with a speed of v' 2.10 m/s. a. Calculate the angle of deflection of particle B relative to the x-axis. b. Calculate the speed of particle B after the collision. c. Calculate the change of kinetic energy in the collision. d. What type of collision is this? Explain. Sou (aExplanation / Answer
Given
two particle of masses mA= m1 = 0.120 kg, vA = u1= 2.80 m/s along the +x direction
mass mB = m2 = 0.140 kg is at rest ==> vB = u2 = 0 m/s
m1 collides with m2 so that m1 deflected off an angle of theta1_f = 30 degrees with a speed of vA' = v1 = 2.10 m/s
unknowns are vB'= v2 =? , theta2_f =?
from conservation of momentum
m1(u1cos theta1_i+u1 sin theta1_i) + m2(u2cos theta2_i+u2 sin theta2_i) = m1(v1cos theta1_f+v1 sin theta1_f) + m2(v2cos theta2_f+v2 sin theta2_f)
here theta1_i = 0 degrees , theta2_i = 0, theta1_f = 30 degrees , theta2_f = ?
substituting the values and equating the components of momentum
0.120(2.80cos0 + 2.80 sin0)+0.14(0 +0) = 0.120(2.10cos30 + 2.10 sin30)+0.14(v2 cos theta2_f +v2sin theta2_f)
0.120 *2.8 cos0 = 0.120(2.10 cos30 )+0.140*v2 cos theta2_f ===> (0.120-0.218) = 0.140*v2 cos theta2_f -----(B)
0 = 0.120( 2.10 sin30) + 0.140*v2 sin theta2_f ====> - 0.126 = 0.140*v2 sin theta2_f ------(A)
dividing the equations (A)/(B)
- 0.126 = 0.140*v2 sin theta2_f
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solving for theta2_f = 52.13 degrees with respect to +x axis
the angle of deflection is theta = 52.13 degrees
b.
substituting the value of theta2_f in any equations we get v2
(A)====> - 0.126 = 0.140*v2 sin 52.13
solving for v2 = -1.1401 m/s
c.
k.e = 0.5*m*v^2
initial kinetic energy is k.e_i = 0.5(mA*vA^2+mB*vB^2) = 0.5(0.12*2.8^2+0.14*(0)^2) J = 0.4704 J
final kinetic energy is k.e_f = 0.5(mA*vA'^2+mB*vB'^2) = 0.5(0.12*2.1^2+0.14*(-1.1401)^2) J = 0.3556 J
d.
so the k.e s are not equal means the collision is inelastic collision