# A 2.0-g particle moving at 8.6

A 2.0-g particle moving at 8.6 m/s makes a perfectly elastichead-on collision with a resting 1.0-g object. (a) Find the speedof each particle after the collision.

2.0 g particle | m/s |

1.0 g particle | m/s |

(b) Find the speed of each particle after the collision if thestationary particle has a mass of 10 g.

2.0 g particle | m/s |

10.0 g particle | m/s |

(c) Find the final kinetic energy of the incident 2.0-g particle inthe situations described in parts (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)

Answer:

m1= mass of ist particle = 2g

u1 = initial velocity of ist particle = 8.6 m/s

u2= initial velocity of 2nd particle = 0 ( due to atrest)

m2 = mass of 2nd particle = 1g

In elastic collision ,velocity of ist particle after collisionis given by:

V1 =[ (m1-m2)*u1 + 2*m2*u2] / (m1+m2)

= ( 2*10^{-3} -1*10^{-3})*8.6 +2*1*10^{-3}*0 / (2+1)*10^{-3}

= 2.867 m/s

V2 = velocity of 2nd particle after collision

V2 = (2*m1*u1 + ( m2-m1)*u2] / (m1+m2)

= 2*2*10^{-3}*8.6 + ( 1*10^{-3} -2*10^{-3})*0 / ( 2+1)*10^{-3}

= 11.47 m/s

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B) now m2 = 10g

V1 = [ (m1-m2)*u1 + 2*m2*u2] / (m1+m2)

= ( 2 – 10)*10^{-3}*8.6 + 2*0 / (2+10)*10^{-3}

= -5.73 m/s

V2 = (2*m1*u1 + ( m2-m1)*u2] / (m1+m2)

= 2*2*10^{-3}*8.6 + ( 10- 2)*10^{-3}*0 / (2+10)*10^{-3}

= 2.867 m/s

—————————————————————————————-

c) final K.E of ist particle in situation a:

K.E = (1/2)*m1*V1^{2}

= (1/2)*2*10^{-3}*( 2.867 m/s)^{2}

= 8.219*10^{-3} J

in situation b:

K.E’ = (1/2)*2*10^{-3}*( -5.73m/s)^{2}

= 0.0328 m/s

in case b particle lose more K.E