question_answer

Fig. 4(HT).3 shows the transverse section of a hill. A ball of mass 0.5 kg is given a K.E. of 300J at A and rolls down to C. Assuming that  there is no friction, calculate (i) K.E. of ball of B (ii) P.E. of ball of C, taking P.E. at A equal to zero (iii) speed of the ball at C; and (iv) change in P.E. in going from B to C. Take \[g=10\,m/{{s}^{2}}\]

Answer:

                (i) Here, m = 0.5 kg K.E.at A = 300 J P.E.at A = 0 In going from A to B, Gain in \[P.E.=mg\left( 39-18 \right)=0.5\times 10\times 21=105J\] (i) As P.E. at A = 0 and C is 18 m below A, therefore, P.E. at C = - mgh = - 0-5 x 10 x 18 = - 90 J (ii) K.E. at C = K.E. at A - P.E. at C = 300 - (- 90) = 390 J From \[\text{K}.\text{E}.\text{ }=\frac{1}{2}\text{m}{{\upsilon }^{\text{2}}}\] \[390=\frac{1}{2}\times 0.5{{\upsilon }^{2}};\,{{\upsilon }^{2}}=\frac{2\times 390}{0.5}=1560\] \[\upsilon =\sqrt{1560}=39.5m{{s}^{-1}}\] (iv) Change in P.E. in going from B to C = P.E. at C - P.E. at B = - 90 - 105 = -195 J