# Solved Examples : Impulse

Illustration : The area of F-t curve is A. If the one of the colliding bodies of mass m is at rest, find its speed just after the collision.

Solution: The area of F-t curve gives the impulse of the force F during the time. That is equal to the change in momentum of the colliding bodies.

Since one of the colliding body was at rest just before the collision, the change in its momentum during the impact is equal to momentum just after the impact, that is mv.

Since , $\displaystyle \int F \delta t = A$

m v = A

⇒ v = A/m.

Illustration : Two block of masses m and 3m are connected by an inextensible string and the string passes over a fixed pulley which is massless and frictionless. A bullet of mass m moving with a velocity ‘ u ‘ hits the hanging block of mass ‘ m ‘ and gets embedded in it. Find the height of through which block A rises after the collision.

Solution: As soon as the collision occurs, the string becomes slack, tension becomes zero. Vertically gravitational force is acting downwards. But as gravitational force is weaker force than impulse force, therefore a possible change of momentum due to gravitational force during collision can be neglected.
Hence, conserving momentum of the system during collision along vertical:

If v = velocity of the combined mass just after collision

Then, mu = 2mv

⇒ v = u/2

Now, maximum height through which the combined mass rises

$\displaystyle H_{max} = \frac{v^2}{2 g} = \frac{u^2}{8 g}$

Illustration : A system of two blocks A and B are connected by an inextensible massless string as shown. The pulley in massless and frictionless. A bullet of mass ‘ m ‘ moving with a velocity ‘ u ‘ as shown hits the block ‘ B ‘ and gets embedded into it. Find the impulse imparted by tension force to the system.

Solution: Let velocity of B and A after collision has magnitude v.

At the time of collision tension = T

Impulse provided by tension = $\displaystyle \int T dt$

Consider change of momentum of (B + bullet)

$\displaystyle m u – \int T dt = 2 m v$ ……… (i)

Consider change of momentum A

$\displaystyle \int T dt = 3 m v$    ………..(ii)

from (i) and (ii), mu = 5 mv

=> v = u/5

Hence, Impulse $\displaystyle \int T dt = 3 m v$

= 3m(u/5) = 3mu/5