Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • An   man is riding on a small  <img alt="40 \mathrm{~kg}" src="https

An 80 \mathrm{~kg}  man is riding on a small  40 \mathrm{~kg}  cart at a speed of  4 \mathrm{~m} / \mathrm{s} . He jumps off the cart with zero horizontal speed. What is the resulting changes in the speed of the cart (in \mathrm{m} / \mathrm{s} )?

Option: 1

4


Option: 2

8


Option: 3

12


Option: 4

0


Answers (1)

best_answer

Apply conservation of momentum to the (man + boat) system.
Total initial momentum, P_i=(80+40) \times 4 \mathrm{~kg} \mathrm{~m} / \mathrm{s}
Total final momentum, P_f=[80 \times 0+40 \times \mathrm{V}]

=40 \mathrm{~V} \mathrm{~kg} \mathrm{~m} / \mathrm{s} .

Now,

P_i=P_f \\
(80+40) \times 4=40 \mathrm{~V} \\
v=12 \mathrm{~m} / \mathrm{s} .
 

Posted by

Ritika Harsh

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET UG 2025 'safe': All state governments on alert, NTA cautions against paper leak, misleading rumours
anam-khan

NEET-UG mock drills underway across centres ahead of May 4 entrance
anam-khan

NEET probe: NMC cancels admission of 14 students, orders suspension of 26 more

Latest Question