A ball of mass m = 1 kg and negligibleideal string, length R

The potential energy P of a particle of mass m at a height h above the reference level is given by, P = mgh where g = 9.8 m/s2 is the acceleration due to gravity. The kinetic energy K of a particle of mass m and speed v is given by, K = 12mv2 The momentum p→ of a particle of mass m and velocity v→ is given by, p→ = mv→ According to the law of conservation of energy, the total mechanical energy (sum of potential and kinetic energy) of an isolated system remains constant. According to the law of conservation of momentum for collisions, the total momentum of the system remains conserved before and after the collision.The mass of the ball is m = 1 kg The length of the string is R = 1 m The ball is dropped from rest therefore the initial speed of the ball is zero The initial height of the ball above the table (reference) is h = R = 1 m The mass of the cube is MThe potential energy of the ball before it is dropped is, P = mgR The kinetic energy of the ball before dropping is zero as the initial speed of the ball is zero. The potential energy of the ball immediately before the collision with the cube is zero (reference level) Let the speed of the ball right before the collision is v. The kinetic energy of the ball immediately before the collision with the cube is, K = 12mv2 Applying the law of conservation of energy for the ball, mgR + 0 = 0 + 12mv2⇒v = 2gR On substituting the known values, v = 2×9.8 m/s2×1 m = 19.6 m2/s2 = 4.43 m/s The speed of the ball immediately before the collision is 4.43 m/s.Before the collision- The ball has a speed v = 4.43 m/s towards the right The cube is at rest so its speed and momentum is zero After the collision, The ball recoils to the left with a speed one-fourth of its speed before the collision, then the speed of the ball after the collision is v2b = v4 The cube recoils to the right with a speed one-fourth of the speed of the ball before the collision, then the velocity of the cube after the collision is v2c = v4 Let the momentum of the ball before and after the collision is p1b and p2b and the momentum of the cube before and after the collision is p1c and p2c then applying the conservation of momentum for the system, p1c + p1b = p2c - p2b Since the ball recoils towards the left, therefore, the negative sign is taken before p2b (momentum is a vector quantity) M0 + mv = Mv2c - mv2b⇒M = mv + v2bv2c⇒M = 1 kgv + v4v4⇒M = 5 kg(A) 4.43 m/s (B) M = 5 kg

Answered: A ball of mass m= 1 kg and negligible…