The velocity of a body of mass 2 kg changes from v1- 2i-3j-k m-s to v2-3i-2j-3 k m-s in 3 sec under the influence of a external force- Find the magnitude of the force applied -

Change in momentum, ∆P=mv2 mv1 nbsp;∆P=m[( 3i^ nbsp;+2j^ nbsp;+3k^) nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbs ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Calculating Friction Using Coeffcients

The velocity ...

Question

The velocity of a body of mass $2 k g$ changes from $v_{1} = 2 ˆ i + 3 ˆ j - ˆ k m / s$ to $v_{2} = - 3 ˆ i + 2 ˆ j + 3 ˆ k m / s$ in $3 s e c$ under the influence of a external force. Find the magnitude of the force applied ?

5.1 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.8 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.3 N

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 4.3 N
$Change in momentum, ΔP = m v_{2} - m v_{1} Δ P = m [(- 3 ˆ i + 2 ˆ j + 3 ˆ k) - (2 ˆ i + 3 ˆ j - ˆ k)] \Rightarrow 2 \times (- 5 ˆ i - ˆ j + 4 ˆ k) = (- 10 ˆ i - ˆ 2 j + 8 ˆ k) (Δ P | = \sqrt{{(- 10)}^{2} + {(- 2)}^{2} + {(8)}^{2}} (Δ P | = 12.96 \frac{k g m}{s} F o r c e, (F | = \frac{(Δ P |}{Δ t} = \frac{12.96}{3} = 4.32 N$

Suggest Corrections

Similar questions

Q. The velocity of a body of mass

2 k g

changes from

v_{1} = 2 ˆ i + 3 ˆ j - ˆ k m / s

v_{2} = - 3 ˆ i + 2 ˆ j + 3 ˆ k m / s

3 s e c

under the influence of a external force. Find the magnitude of the force applied ?

A constant force acts on a body of mass 5kgfor a duration of 2s. It increases the object's velocity from 3m/s to 7m/s . Find the magnitude of the force that was applied for a duration of 5s, what would be the final velocity of the object?

Q. The velocity of the centre of mass of a system changes from

_{1} = 4^i m/s

_{2} = 3^j m/s

during time

Δ t = 2 s

. If the mass of the system is

m = 10 kg

, the magnitude of constant force acting on the system is

Q. A force is applied to box of mass 4 kg and it changes the velocity from 3 m/s to 6 m/s in 8 s. Determine the work done by force during the this process.

Q. A constant force

F

is applied on a body of mass

5 k g

, changes its velocity from

2 m / s

7 m / s

10 s

. If the direction of motion is unchanged, then the

F

is :

Join BYJU'S Learning Program

The velocity of a body of mass 2 kg changes from v1= 2ˆi+3ˆj−ˆk m/s to v2=−3ˆi+2ˆj+3 ˆk m/s in 3 sec under the influence of a external force. Find the magnitude of the force applied ?

The correct option is C 4.3 N Change in momentum, ΔP=mv2−mv1 ΔP=m[(−3ˆi +2ˆj +3ˆk) −(2ˆi +3ˆj −ˆk)] ⇒ 2×(−5ˆi −ˆj +4ˆk) = (−10ˆi −ˆ2j +8ˆk)(ΔP|=√(−10)2 +(−2)2+(8)2 (ΔP|=12.96 kg m/s Force, (F|=(ΔP|Δt=12.963=4.32 N

The velocity of a body of mass $2 k g$ changes from $v_{1} = 2 ˆ i + 3 ˆ j - ˆ k m / s$ to $v_{2} = - 3 ˆ i + 2 ˆ j + 3 ˆ k m / s$ in $3 s e c$ under the influence of a external force. Find the magnitude of the force applied ?