Suppose a 0.250-kg ball is thrown at 15.0 m/s to a motionless person standing on ice who catches it with an outstretched arm as shown in Figure 10.40. (a) Calculate the final linear velocity of the person, given his mass is 70.0 kg. (b) What is his angular velocity if each arm is 5.00 kg? You may treat the ball as a point mass and treat the person's arms as uniform rods (each has a length of 0.900 m) and the rest of his body as a uniform cylinder of radius 0.180 m. Neglect the effect of the ball on his center of mass so that his center of mass remains in his geometrical center. (c) Compare the initial and final total kinetic energies.

Step-by-step solution Step 1 of 8 There is no external force is involved in the system, so from the conservation of the linear momentum. Here, is the mass of the ball, is the linear velocity of the ball before the collision, is the mass of the man and is the final linear velocity. Step 2 of 8 (a) Substitute 0.250 kg for , 15.0 m/s for and 70.0 kg for . . Hence, the required final linear velocity of the person is . Step 3 of 8 (b) The arms are assumed as the rods, so its moment of inertia about the body axis of rotation is, Here, m is the mass of a single arm and l is the length of the arm. Step 4 of 8 The moment of inertia of its body is, Here, M is the mass of the body (without hands) and r is the radius of body which is assumed as cylindrical shaped. Substitute 60.0 kg for M and 0.180 m for r. Total moment of inertia is calculated as,