The winch on the back of the jeep A is turned on and pulls the tow rope at speed 2m/s.The mass of jeep A is 2.5×103kg while that of car B is 1.25×103kg . The length of the rope is 5m. Select the correct alternative.

Speeds of A and B will be shown.By conservation of momentum, 2.5×103×v=1.25×103×(2−v)⇒v=23m/s=speed of Aand speed of B=2−23=43m/sAs relative speed is 2m/s, time after which A and B meet is 52=2.5s

The winch on the back of the jeep A is turned on and pulls the tow rope at speed 2m/s.The mass of jeep A is 2.5×103kg while that of car B is 1.25×103kg . The length of the rope is 5m. Select the correct alternative.

Speeds of A and B will be shown.By conservation of momentum, 2.5×103×v=1.25×103×(2−v)⇒v=23m/s=speed of Aand speed of B=2−23=43m/sAs relative speed is 2m/s, time after which A and B meet is 52=2.5s

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The winch on the back of the jeep A is turned on and pulls the tow rope at speed $2 m / s .$ The mass of jeep A is $2.5 \times 10^{3} k g$ while that of car B is $1.25 \times 10^{3} k g$ . The length of the rope is $5 m .$ Select the correct alternative.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

When they meet, speed of A is $0.67 m s^{- 1}$

When they meet, speed of B is $1.33 m s^{- 1}$

They will meet after a time $2.50 s$

Speed of A will be zero, while that of B will be $2 m / s$

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Speeds of A and B will be shown.
By conservation of momentum,
Question Image
$\begin{array}{l} 2.5 \times 10^{3} \times v = 1.25 \times 10^{3} \times (2 - v) \\ \Rightarrow v = \frac{2}{3} m / s = s p e e d o f A \\ a n d s p e e d o f B = 2 - \frac{2}{3} = \frac{4}{3} m / s \end{array}$
As relative speed is $2 m / s$ , time after which A and B meet is $\frac{5}{2} = 2.5 s$