3. (24) A 2 kg block on a frictionless horizontal track is initially moving to t

Question

3. (24) A 2 kg block on a frictionless horizontal track is initially moving to the right at 20 m/s. At t=0 it collides with an 8 kg block that is initially stationary. Assume that this is an elastic collision.

a) Before the collision, what is the kinetic energy and the momentum of the 2 kg block?

b) After the collision, what are the speeds of the blocks?

c) What is the kinetic energy of the 2 kg block after the collision?

d) What is the kinetic energy of the 8 kg block after the collision?

e) Plot the velocity of the 2 kg block as a function of time from t = -2 s to t = +2 s.

f) Plot the kinetic energy of the 2 kg block vs t from -2 to +2 s. On the same graph plot the kinetic energy of the 8 kg block from -2 to +2 s. What do these plots show?

Explanation / Answer

a) Before collision the kinetic energy of the system is

K0 = K10 + K02   

K0 = ( m1 v102 ) / 2 + ( m2 v202 ) / 2

K0 = ( 2.0 kg * ( 20 m/s )2 ) / 2 + ( 8.0 kg * ( 0.0 m/s )2 ) / 2

K0 = 400 J

b) before collision

P0 = m1 v10

after collision

P = m1 v1 + m2 v2

K = ( m1 v12 ) / 2 + ( m2 v22 ) / 2

the, considering the conservation of the linear momentum

P = P0

m1 v1 + m2 v2 = m1 v10   --->   v2 = [ m1 ( v10 - v1 ) ] / m2

since collision is elastic, you can consider the conservation of the kinetic energy

K = K0

( m1 v12 ) / 2 + ( m2 v22 ) / 2 = K0

m1 v12 + m2 v22 = 2K0

m1 v12 + m2 ( [ m1 ( v10 - v1 ) ] / m2 )2 = 2K0

m1 v12 + [ m1 ( v10 - v1 ) ]2 / m2  = 2K0

m1 m2 v12 + m12( v102 - 2 v10 v1 + v12 ) = 2 m2 K0

m1 m2 v12 + m12 v102 - 2 m12 v10 v1 + m12 v12 - 2 m2 K0 = 0

[ m1 m2 + m12 ] v12  - 2 m12 v10 v1 + [ m12 v102 - 2 m2 K0 ] = 0

[ ( 2 kg ) ( 8 kg )+ ( 2 kg )2 ] v12  - 2 ( 2 kg )2 ( 20 m/s ) v1 + [ ( 2 kg )2 ( 20 m/s )2 - 2 ( 8 kg ) ( 400 J ) ] = 0

( 20 kg2 ) v12 - ( 160 kg2m/s ) v1 - ( 4800 kg2m2/s2 ) = 0

v12 - ( 8 m/s ) v1 - ( 240 m2/s2 ) = 0 ---> v1 = { -12 m/s , 20 m/s}

we chose the value v1 = -12 m/s

then,

v2 = [ 2 Kg *( 20 m/s- ( -12 m/s ) ) ] / ( 8 Kg )

v2 = 8 m/s

c) the kinetic energy of the block 8 Kg is

K2 = ( m2 v22 ) / 2

K2 = ( 8 Kg * ( 8 m/s )2 ) / 2

K2 = 256 J

d) the kinetic energy of the block 2 Kg is ,,,,,,,,, K1 = ( m1 v12 ) / 2 = ( 2 Kg * ( 12 m/s )2 ) / 2 = 144 J

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