Two masses, `2m` kg and `4m` kg, are attached by a light string. The string is placed over a smooth pulley as shown.
The two masses are at rest before being released and `v` is the velocity of the larger mass at time `t` seconds after they are released.
The force due to air resistance on each mass has magnitude `kv`, where `k` is a positive constant.
- Show that `frac{dv}{dt} = frac{gm-kv}{3m}`. (2 marks)
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- Given that `v < frac{gm}{k}`, show that when `t = frac{3m}{k} ln 2`, the velocity of the larger mass is `frac{gm}{2k}`. (3 marks)
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i.
`text{Taking} \ v \ text{downwards as positive.}`
`text{Forces acting on}\ 2m\ text{mass:}`
`kv + 2 mg-T = -2m * frac{dv}{dt}\ …\ (1)`
`text{Forces acting our} \ 4m \ text{mass:}`
`4mg-kv-T = 4m * frac{dv}{dt}\ …\ (2)`
`text{Subtract} \ \ (2)-(1)`
| `2 mg-2 kv` | `= 6 m * frac{dv}{dt}` |
| `:. frac{dv}{dt}` | `= frac{2mg-2 kv}{6 m}` |
| `= frac{gm-kv}{3m}` |
| ii. | `frac{dv}{dt}` | `= frac{gm-kv}{3m}` |
| `frac{dt}{dv}` | `= frac{3m}{gm-kv}` | |
| `t` | `= int frac{3m}{gm-kv}\ dv` | |
| `= -frac{3m}{k} log_e |gm-kv | + c` |
`text{When} \ \ t = 0, v = 0:`
| `0` | `= -frac{3m}{k} log_e \ | gm | + c` |
| `c` | `= frac{3m}{k} log_e \ | gm | ` |
| `t` | `= frac{3m}{k} log_e \ | gm | \-frac{3m}{k} log_e \ | gm -kv |` |
| `= frac{3m}{k} log_e \ | frac{mg}{gm-kv} |` |
`text{Find} \ \ v\ \ text{when} \ \ t = frac{3m}{k} log_e 2 :`
| `frac{3m}{k} log_e 2` | `= frac{3m}{k} log_e | frac{gm}{gm-kv} |` |
| `2` | `= frac{gm}{gm-kv}` |
| `2gm-2kv` | `= gm` |
| `2kv` | `= gm` |
| `therefore \ v` | `= frac{gm}{2k}` |