A mass of 3 kg descending vertically downward supports a mass of 2 kg by means the end of 5 s, the string breaks. How much higher the 2 kg mass will go further?

Question

A mass of 3 kg descending vertically downward supports a mass of 2 kg by means the end of 5 s, the string breaks. How much higher the 2 kg mass will go further? (A) 4.9 m (B) 9.8 m (C) 19.6 m (D) 2.45 m

Tardigrade Admin · Accepted Answer

Acceleration of combined system, a=(m1-m2/m1+m2) ⋅ g=(3-2/3+2) × 9.8=1.96 ms -2 Vertically upward velocity of 2 kg mass at the time of breaking of string, v=a t=5 × 1.96=9.8 ms -2. After breaking of string, mass m2 moves under gravity and go further higher through a height h, where final velocity is zero. Hence, (0)2-(9.8)2 =2 ×(-9.8) × h or h =4.9 m

Q. A mass of 3kg descending vertically downward supports a mass of 2kg by means the end of 5s, the string breaks. How much higher the 2kg mass will go further?

4.9 m

9.8 m

19.6 m

2.45 m

Q. A mass of $3 k g$ descending vertically downward supports a mass of $2 k g$ by means the end of $5 s$ , the string breaks. How much higher the $2 k g$ mass will go further?