A 2.00-f particle moving at 8.20 makes a perfecly elastic head-on collision with

Question

A 2.00-f particle moving at 8.20 makes a perfecly elastic head-on collision with a resting 1.00-g object. (Assume the 2.00-g particle is moving in the positive direction before the collision. Indicate the direction with the sign of your answer.) (a) Find the velocity of each particle after the collision. 2.00-g particle m/s 1.00-g particle m/s (b) Find the velocity of each particle after the collision if the stationary particle has a mass of 10.0 g. 2.00-g particle m/s 10.0-g particle m/s (c) Find the final kinetic energy of the incident 2.00-g particle in the situations described in pats (a) and (b). KE in part (a) J KE in part (b) J In which case does the incident particle lose more kinetic energy? case (a) case (b)

Explanation / Answer

given data

m1 = 2 g, m2=1 g , speed of the m1, u1= 8.2 m/s

speed of m2, u2 = 0

from the conservation of linear momentum

m1*u1 + m2*u2 = m1*v1 + m2*v2

Likewise, the conservation of the total kinetic energy is expressed by the equation

1/2*m1*u1² + 1/2*m2*u2² = 1/2*m1*v1² +1/*m2*v2²

These two equations may be solved directly to find vi when ui are known or vice versa.

v1 =(u1*(m1 - m2) + 2*m2*u2)/(m1 +m2) and v2 =(u2*(m2 - m1) + 2*m1*u1)/(m1 + m2)

Than v1 = (8.2*(2 - 1) + 2*1*0)/3=2.733m/s and

v2 =(0*1 + 2*2*8.2)/3 = 9.6 m/s

b)Than v1 = (8.2*(2 - 10) + 2*1*0)/12= 5.466m/s

v2 =(0*1 + 2*2*8.2)/12 = 4/3*u1 = 2.73 m/s

C)

in case a)

v1 = 2.733 m/s

KE=1/2*m*v1^2 = 0.5*2*10^-3*2.733*2.733 =7.469 *10^-3 =0.007469

incase of b) v1=5.466

KE =1/2*m*v1^2 =0.5*2*10^-3*2.733*5.466*5.466 =81.65*10^-3 =0.08165

In case a, it has smaller KE

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