JPJC 2026 Quantities and Measurement Assignment Solutions

1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS/8867 H1 PHYSICS QUANTITIES AND MEASUREMENT ASSIGNMENT SOLUTIONS Qn. Solution Mark Allocation 1 12 12 (2 ) 2 o o L FdF dL  =  =II II 2 22(kg m s )(m)Units of kg m s A(A)(A)(m) o − −−== [2] Deduct 1 mark for each incorrect units of 12,II , F, d and L, up to max. of 2 marks. [1] for correct units of o 2 SI base units of 23kg m sP −= Base units of ( ) ( ) ( ) 23kg m s m kg m wy xw x ykF L  −−= 32kg m sw y w x y w+ + − −= Comparing indices of similar units on both sides of the equation, w = 1.5, x = −1, y = −0.5 [2] Deduct 1 mark for each incorrect units of P, F, L and  , up to max. of 2 marks. [1] for correct comparing of indices. [1] for correct answer. 3(a) Component of resultant force along OX 4.0cos30 5.0cos60 0.9641 0.96 N = −  =  [1] for resolving forces correctly into their components 3(b) Component of resultant force along OY 4.0sin30 5.0sin60 3.0 3.330 3.3 N = + − =  Resultant force ( ) ( ) 22 0.9641 3.330 3.4668 3.5 N =+ =  3.330tan 0.9641 = 73.9 74 =   Therefore, the resultant force is 3.5 N at an angle of 74° above the x-axis. [1] for resolving forces correctly into their components [1] for correct calculation of resultant force [1] for correct calculation of direction