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

Byju's Answer

Standard XII

Physics

Intro to Collision

Two point mas...

Question

Two point masses connected by an ideal string are placed on a smooth horizontal surfaces as shown in the diagram. A sharp impulse of $10 kg-m/s$ is given to the $5 kg$ mass. The velocity of the $10 kg$ mass just after application of impulse will be $\frac{1}{x} m/s$ , then value of $x$ is

Open in App

Solution

$10 kg$ mass will gain velocity along the string only. Using momentum conservation along the string
$v = \frac{10 cos 60^{\circ}}{10 + 5} = \frac{1}{3} m/s$

Suggest Corrections

Similar questions

Q. Two point masses connected by an ideal string are placed on a smooth horizontal surfaces as shown in the diagram. A sharp impulse of

10 kg-m/s

is given to the

5 kg

mass. The velocity of the

10 kg

mass just after application of impulse will be

\frac{1}{x} m/s

, then value of

x

Q. Two point masses connected by an ideal string are placed on a smooth horizontal surface as shown in the diagram. A sharp impulse of 10

k g m l s

is given to the 5

k g

mass. The velocity of the 10

k g

mass just after the impulse, will be

Q. Two point masses connected by a light inextensible string are lying on a frictionless surface as shown in figure. An impulse of magnitude

10 kg m/s

is given to the

5 kg

block. Then:

Q. Two blocks of masses

10 k g

and

4 k g

are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of

14 m / s

to the heavier block in the direction of the lighter block, The velocity of the centre of mass is:

Two blocks of the masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block. The velocity of centre of mass is

Join BYJU'S Learning Program

Two point masses connected by an ideal string are placed on a smooth horizontal surfaces as shown in the diagram. A sharp impulse of 10 kg-m/s is given to the 5 kg mass. The velocity of the 10 kg mass just after application of impulse will be 1x m/s, then value of x is

10 kg mass will gain velocity along the string only. Using momentum conservation along the string v=10cos60∘10+5=13 m/s

Two point masses connected by an ideal string are placed on a smooth horizontal surfaces as shown in the diagram. A sharp impulse of $10 kg-m/s$ is given to the $5 kg$ mass. The velocity of the $10 kg$ mass just after application of impulse will be $\frac{1}{x} m/s$ , then value of $x$ is

$10 kg$ mass will gain velocity along the string only. Using momentum conservation along the string
$v = \frac{10 cos 60^{\circ}}{10 + 5} = \frac{1}{3} m/s$