ILLUSTRATION- A particle a mud pallet- sayof mass m strikes a smooth stationary wedge of mass M with a velocity v 0- at an angle - with horizontal- If the collision is perfectly inelastic- find the velocity of the wedge just after the collision-A- mv 0sin- m - MB- m v0tan-m-MC- m v0cos-m-MD- Mv 0cos- m - M

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Standard XII

Physics

Perfectly Inelastic Collision

ILLUSTRATION:...

Question

ILLUSTRATION: A particle (a mud pallet, say)of mass m strikes a smooth stationary wedge of mass $M$

with a velocity $v_{0}$ , at an angle $θ$ with horizontal. If the collision is perfectly inelastic, find the velocity of

the wedge just after the collision.

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Solution

The correct option is C

Let the system $(M + m)$ moves as a single mass with a velocity $v$ .

Conserving the momentum of the system in horizontal, we have $m v_{0} c o s θ = (M + m) v$

This gives $v = \frac{m v_{0} c o s θ}{M + m}$

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Similar questions

A particle (a mud pallet) of mass 'm' strikes a smooth stationary wedge of mass $M$ with a velocity $v_{0}$ , at an angle $θ$ with horizontal. If the collision is perfectly inelastic, find the velocity of the wedge just after the collision.

Q. A particle (a mud pallet, say) of mass

m

strikes a smooth stationary wedge of mass

M

with a velocity

ν_{0}

, at an angle

θ

with horizontal. If the collision is perfectly inelastic, find the:
a) velocity of the wedge just after the collision.
b) change in

K E

of the system

(M + m)

in collision.

Q. A particle of mass 'm' strikes a smooth stationary wedge of mass M with a velocity

v_{0}

, at an angle

θ

with horizontal if the collision is perfectly inelastic, the impulse on the wedge is:

A particle of mass 'm' strikes a wedge of mass 'M' horizontally as shown in the figure.
STATEMENT-1 : If the collision is perfectly inelastic then, it can be concluded that the particle sticks to the wedge.
STATEMENT-2 : In perfectly inelastic collision velocity of both bodies is same along common normal just after the collision.

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ILLUSTRATION: A particle (a mud pallet, say)of mass m strikes a smooth stationary wedge of mass M with a velocity v0, at an angle θ with horizontal. If the collision is perfectly inelastic, find the velocity of the wedge just after the collision.

The correct option is C Let the system (M + m) moves as a single mass with a velocity v. Conserving the momentum of the system in horizontal, we have mv0cos θ = (M + m)v This gives v = mv0cos θM + m

ILLUSTRATION: A particle (a mud pallet, say)of mass m strikes a smooth stationary wedge of mass $M$

with a velocity $v_{0}$ , at an angle $θ$ with horizontal. If the collision is perfectly inelastic, find the velocity of

the wedge just after the collision.

The correct option is C

Let the system $(M + m)$ moves as a single mass with a velocity $v$ .

Conserving the momentum of the system in horizontal, we have $m v_{0} c o s θ = (M + m) v$

This gives $v = \frac{m v_{0} c o s θ}{M + m}$