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A particle is given velocity `v_0` at point A as shown in the figure. Find the speed of particle at point `B,C` and `D` |
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Answer» `K_A=(1)/(2)mv_0^2`,`U_A=mg(2h)=2mgh` `K_B=(1)/(2)mv_1^2`,`U_B=mgh` `K_C=(1)/(2)mv_2^2`,`U_C=-mg(2h)=-2mgh` `K_D=(1)/(2)mv_3^2`,`U_D=0` `K_A+U_A=K_B+U_B` `implies(1)/(2)mv_0^2+2mgh=(1)/(2)mg^2+mgh` `impliesv_1=sqrt(v_0^2+2gh)` `=K_C+U_Cimpliesv_2=sqrt(v_0^2+8gh)` `=K_D+U_Dimpliesv_3=sqrt(v_0^2+4gh)` |
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