A 0.500 kg sphere moving with a velocity (2.00i - 2.90j + 1.00k) m/s strikes ano

Question

A 0.500 kg sphere moving with a velocity (2.00i - 2.90j + 1.00k) m/s strikes another sphere of mass 1.50 kg moving with a velocity (-1.00i + 2.00j - 2.50k) m/s.

(a) If the velocity of the 0.500 kg sphere after the collision is (-0.90i + 3j - 8.00k) m/s, find the velocity of the 1.50 kg sphere and identify the kind of collision (elastic, inelastic, or perfectly inelastic).

(b) If the velocity of the 0.500 kg sphere after the collision is (-0.250i + 0.775j - 1.63k) m/s, find the final velocity of the 1.50 kg sphere and identify the kind of collision.

(c) What if? If the velocity of the 0.500 kg sphere after the collision is (-1.00i + 3.10j + ak) m/s, find the value of a and the velocity of the 1.50 kg sphere after an elastic collision. (Two values of a are possible, a negative value and a positive value. Report each with their corresponding final velocities.)

Explanation / Answer

The question deals with the law of conservation of momentum and energy.
Let us name the following way,
0.500 kg sphere, mass, m1 = 0.500kg
initial velocity, v1 = (2.00i - 2.90j + 1.00k) m/s
final velocity = v1'
1.500 kg sphere, mass, m2 = 1.500kg
initial velocity, v2 = (-1.00i + 2.00j - 2.50k) m/s.
final velocity = v2'
Law of conservation of momentum states, m1*v1 +m2*v2 = m1*v1'+m2*v2'
and to find the kind of collision we use the kinetic energies involved,
for elastic collisions, 0.5*m1*(v1^2) +0.5*m2*(v2^2) = 0.5*m1*(v1'^2)+0.5*m2*(v2'^2)
for inelastic collisions, 0.5*m1*(v1^2) +0.5*m2*(v2^2) > 0.5*m1*(v1'^2)+0.5*m2*(v2'^2) > 0
for perfectly inelastic collisions, 0.5*m1*(v1'^2)+0.5*m2*(v2'^2) = 0
for vectors, v^2 = v.v (dot product)

(a) given v1' =(-0.90i + 3j - 8.00k) m/s
Applying the law of conservation of momentum,
(1i-1.45j+0.5k)+(-1.5i+3j-3.75k)=(-0.45i+1.5j-4k)+m2*v2'
or, 1.5*v2' = (1i-1.45j+0.5k)+(-1.5i+3j-3.75k)-(-0.45i+1.5j-4k)
or, v2' = (-0.0333i+0.0333j+0.5k) m/s
Calculating initial kinetic energy = 0.5*m1*(v1^2) +0.5*m2*(v2^2)
= 0.5*0.5*(4+8.41+1)+0.5*1.5*(1+4+6.25)
= 11.79J
Calculating final kinetic energy = 0.5*m1*(v1'^2)+0.5*m2*(v2'^2)
(There seems to be a typing error in the question since the final kinetic energy of m1 is more than the initial kinetic energy of both the masses. Based on just given data.)

(b) given v1' = (-0.250i + 0.775j - 1.63k) m/s
Applying the law of conservation of momentum,
(1i-1.45j+0.5k)+(-1.5i+3j-3.75k)=(-0.125i+0.3875j-0.815k)+m2*v2'
or, 1.5*v2' = (1i-1.45j+0.5k)+(-1.5i+3j-3.75k)-(-0.125i+0.3875j-0.815k)
or, v2' = (-0.25i+1.775j-1.62333k) m/s
Calculating initial kinetic energy = 0.5*m1*(v1^2) +0.5*m2*(v2^2)
= 0.5*0.5*(4+8.41+1)+0.5*1.5*(1+4+6.25)
= 11.79J
Calculating final kinetic energy = 0.5*m1*(v1'^2)+0.5*m2*(v2'^2)
= 5.21J
So collision is inelastic.

(c)given v1' =(-1.00i + 3.10j + ak) m/s
Applying the law of conservation of momentum,
(1i-1.45j+0.5k)+(-1.5i+3j-3.75k)=(-0.5i+1.55j+0.5ak)+m2*v2'
or, m2*v2' = (1i-1.45j+0.5k)+(-1.5i+3j-3.75k)-(-0.5i+1.55j+0.5ak)
or, v2' = {(-0.33333a-2.17)k} m/s
Calculating initial kinetic energy = 0.5*m1*(v1^2) +0.5*m2*(v2^2)
= 0.5*0.5*(4+8.41+1)+0.5*1.5*(1+4+6.25)
= 11.79J
For elastic collisions, initial kinetic energy = final kinetic energy
hence, 0.5*m1*(v1'^2)+0.5*m2*(v2'^2)=11.79
or, 2.6525+0.25a^2-0.25a-1.6275=11.79
or, 0.25a^2-0.25a-10.765=0
or, a^2-a-43.06=0
hence, a = 7.081 or -6.081

if a = 7.081,
v1' = (-1.00i + 3.10j + 7.081k) m/s
v2' = (-4.53k) m/s

if a = -6.081,
v1' = (-1.00i + 3.10j - 6.081k) m/s
v2' = (-0.143k) m/s

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