A steel ball M has a mass of 1.5 kg and is moving on a frictionless horizontal s

Question

A steel ball M has a mass of 1.5 kg and is moving on a frictionless horizontal surtace with a velocity of 5 m/s. As it rolls, it makes a direct collision with a steel ball N which has a mass of 2.0 kg and is originally at rest. I the coefficient of restitution is 0.6, determine the energy lost on impact. [30 pts] BallA has a mass of 250 g and an initial velocity of 2 m/s. As it rolls on a horizontal frictionless plane, it makes a direct collision with Ball B, which has a mass of 200 g and is originally at rest. If 1, Determine the velocity of each ball after the collision. show that the kinetic energy is conserved. [20 pts]

Explanation / Answer

Mass of steel ball, M = 1.5 Kg

Initial velocity of the steel ball, M = u1 = 5 m/s

Mass of steel ball, N = 2.0 Kg

Initial velocity of the steel ball, N = 0 m/s

Let v1 and v2 be the final velocities of balls M and N, respectively.

According to law of conservation of momentum,

Mu1 + Nu2 = Mv­1 + Nv2

So, 1.5 * 5 + 2.0 * 0 = 1.5v1 + 2.0v2

Thus, 7.5 = 1.5 v1 + 2.0 v2 -----------(1)

Coefficient of restitution, e = (v2-v1)/(u2-u1) = 0.6

e = (v2-v1)/(0-5) = 0.6

So, v2-v1 = -3

Thus, v2 = v1-3 ----------(2)

Solving equations (1) & (2), we get v1 = 3.85 m/s and v2 = 0.85 m/s

Energy before collision = ½ Mu12 + ½ Nu22

= ½ * 1.5*25 + ½ * 2.0* 0

                                        = 18.75 j

Energy after collision = ½ Mv12 + ½ Nv22

= ½ * 1.5*3.852 + ½ * 2.0* 0.852

                                        = 14.43 J

Loss of energy due to impact = 18.75 – 14.43 = 4.31 J

Mass of ball A, MA = 250 gm = 0.25 Kg

Initial velocity of the ball A = u1 = 2 m/s

Mass of ball B, MB = 200 gm = 0.2 Kg

Initial velocity of the ball B = u2 = 0 m/s

Let v1 and v2 be the final velocities of balls A and B, respectively.

According to law of conservation of momentum,

MAu1 + MBu2 = MAv­1 + MBv2

So, 0.25 * 2 + 0.2* 0 = 0.25*v1 + 0.2*v2

Thus, 0.5 = 0.25 v1 + 0.2 v2 -----------(1)

Coefficient of restitution, e = (v2-v1)/(u2-u1) = 1

                                              e = (v2-v1)/(0-2) = 1

So, v2-v1 = -2

Thus, v2 = v1-2 ----------(2)

Solving equations (1) & (2), we get v1 = 2 m/s and v2 = 0 m/s

Energy before collision = ½ MA u12 + ½ MB u22

                                         = ½ * 0.25*4 + ½ * 0.2* 0

                                        = 0.5 J

Energy after collision = ½ MA v12 + ½ MB v22

                                         = ½ * 0.25*4 + ½ *0.2*0

                                        = 0.5 J

Since K.E. before collision is equal to k.E. after collision, energy is conserved.

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