Saved Bookmarks
| 1. |
A cat is chasinga mouseas shown in the figure . The mouse runs horizontally with speed 19m/s. The catruns with a constnatspeed of 20m/s. At whichangleto thehorizontal should the cat run in orderto catch the mouse ? |
|
Answer» Solution :Method -I [Analysis of velocitycomponents] Assumethat the cat and the mousedo meet at POINT D as shown. Let D be the horizontal pointaboveD. Thecat has velcoity.v. which carries it to point D in time t. Sincethe velocityis inclined , we can .D.. The cat hasa velcoities `v_(x),v_(y),v_(A)` carries cat from POINTC to Din timet and `v_(y)` carries cat from Cto Dint . Wethussee that the X-component of velocity`v_(x)` is responsiblefor keepingcat directly above mouse .  Y -component of velocitybrings cat nearnearrer tomouse . Hencethe catshould run such that `v_(x) = 10` . `therefore20 cos alpha = 10` . `cos alpha =1//2, alpha = 60^(@)`. Only when cat runsat this angleit will reamin at thesame level as themouseat all time . To FIND the timeat whichthey meet , we analyze motion alongY - axis . The cat has a Y - component ofvelocityof 20 sin `60 =10 sqrt(13) = 17.32 m//s.` It HASTO travel a distanceof173 m in -ve Y direction. Hence time taken ` t = ("vertical distance ")/("vertical component velcoity") = (173)/(17.3) = 10 sec`. |
|