Question

List 2 shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown. List 1 gives some statements about X and/or Y. Match these statements to the appropriate system(s) from List 2.

Answer 1

1. constant velocity, thus net force=0, $F_X=Mg$

Since X is fixed, PE does not change.

Since Y loses PE (KE constant), ME reduces

Since weight does not pass through P, $\tau \ne 0$.

2.Force on Y by $Z=mg$ downwards. Thus, $F_X=Mg$

Since going up, ${PE}_X$ increases, ME increases

Since weight through center, $\tau =0$

3.$F_X=$ resultant of tensions (mg each) and $mg\ne Mg$. Going down, PE decreases, ME decreases.

Weight does not pass through P, $\tau \ne 0$

4.a decreases, thus $F_x<Mg$

As Y decreases, X increases, so PE increases. Isolated system, thus ME=constant and $\tau \ne 0$.

5. a=0, thus $F_x=Mg$.

As Y decreases, X increases, so PE increases. Due to viscous force, ME decreases, $\tau \ne 0$

Answer is A$\to$(1,5) B$\to$(2,4,5) C$\to$(1,3,5) D$\to$(2).