A 7.0-kg shell at rest explodes into two fragments, one with a mass of 2.0 kg and the other with a mass of 5.0 kg. If the heavier fragment gains 100 J of kinetic energy from the explosion, how much kinetic energy does the lighter one gain?

The gain in the kinetic energy of the smaller fragment is 249.64 J.

The given parameters;

The velocity of the heavier fragment is calculated as follows;

Apply the principle of conservation of linear momentum to determine the velocity of the smaller fragment as;

The gain in the kinetic energy of the smaller fragment is calculated as follows;

Thus, the gain in the kinetic energy of the smaller fragment is 249.64 J.

Learn more about conservation of linear momentum here: link

Explanation:

Kinetic energy is expressed as KE = 1/2 mv² where;

m is the mass of the body

v is the velocity of the body.

For the heavier shell;

m = 5kg

KE gained = 100J

Substituting this values into the formula above to get the velocity v;

100 = 1/2 * 5 * v²

5v² = 200

v² = 200/5

v² = 40

v = √40

v = 6.32 m/s

Note that after the explosion, both body fragments will possess the same velocity.

For the lighter shell;

mass = 2.0kg and v = 6.32m/s

KE of the lighter shell = 1/2 * 2 * 6.32²

KE of the lighter shell = 6.32²

KE of the lighter shell= 39.94m/s

Hence, the lighter one gains a kinetic energy of 39.94m/s.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s