DHS 04 Forces (Tutorial Solutions)

DUNMAN HIGH SCHOOL 9749 H2/ 8867 H1 PHYSICS Year 5(2024) 1 1. C 2. B 3. (a) When the boat is in equilibrium, By Principle of Floatation: U = msg = 1.2 ×107 (9.81) = 1.18 ×108 N (b) By Archimedes’s Principle: U = ρwVdisplacedg = 1.18 ×108 (1000)Vdisplaced(9.81) = 1.18 ×108 Vdisplaced = 1.20 ×104 m3 Volume of boat below the water line = Vdisplaced = 1.20 × 104 m3 4. (a) kA = 6.0 N m-1 kB = 3.0 N m-1 Tension in A and B = 0.6 N extension in spring A = (0.6 / 6) = 0.10 m extension in spring B = (0.6 / 3) = 0.20 m (b) Total extension = 0.20 + 0.10 = 0.30 m Hence, the effective spring constant of A and B = (0.60 / 0.3) = 2.0 N m-1 5. Acceleration of both trolleys = 0 . 40 . 5 20 = m s-2 Consider the 3.0 kg trolley: 120 . 4 0 . 3 '= × =F N Compression of spring = 0 . 30 . 4 12 = cm Ans: B 3.0 kg force by spring, F’

2 6. Method 1 Let the spring constant of each spring be k. In the first case, the downward force on each spring is W/3. So, the spring constant is given by: kxW =3 ⇒ x Wk 3= In the second case, the force on each spring is (2W)/2 = W Extension e is given by: keW = ⇒ xx W W k We 3) 3/( = = = Method 2 Let the spring constant of each spring be k. Using the formula for parallel springs, In the first case, k eff = k1 + k2 + k3 = k + k + k = 3k Using Hooke’s Law, “F 1 = k1x1” W = 3k x ---------------- (1) In the second case, k eff = k1 + k2 = k + k = 2k Using Hooke’s Law, “F 2 = k2x2” 2W = 2k x2 ---------------- (2) Sub (1) into (2) 2(3k x) = 2k x2 x2 = 3x

3 7. Method 1 Let k be the spring constant of the spring. By Hooke’s Law, 1 2 m N 100010 0 . 2 20 − − =×= =x Fk Elastic potential energy (PE) stored in the spring when it is compressed by 3.0 cm: 45 . 0 ) 030 . 0 )( 1000 (2 1 2 1 22 == =kxU J When the toy reaches its maximum height h above its point of release, Gain in gravitational PE = Loss in elastic PE mgh = 0.45 J ⇒ h = 0.917 m Position 1 Position 2 Position 3 (At max compression) (when toy just leaves spring) (At max height) Method 2 Energy conversion from Position 1 to Position 2 Gain in GPE + Gain in KE = Loss in EPE mgx + ½ mu2 = ½ kx2 (0.050) (9.81) (0.030) + ½ (0.050)u2 = ½ (1000)(0.030)2 u = 4.17 m s-1 (upwards) where u is the velocity of the toy which just leaves the spring Using equation of motion between Position 2 and Position 3, Taking upwards as positive, v 2 = u2 + 2as 02 = (4.17)2 + 2(-9.81)s s = 0.886 m h = x + s = 0.030 + 0.886 = 0.916 m x = 0.030 m s h

4 8. (a) When the block is submerged in water, there is a pressure difference between