A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is

Question

A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is (A) MV (B) 1.5 MV (C) 2 MV (D) zero

Tardigrade Admin · Accepted Answer

A Body of mass =m velocity =V( hati) Knock out (Bounce velocity) =V(- hati) Change in momentum =m(v)-(m)(-v) =2 M V Rate of change of momentum is Force Thus (d vecp/d t)= force. force × time = Impulse Thus Impulse Experience by body =2 m v .

Q. A body of mass $M$ hits normally a rigid wall with velocity $V$ and bounces back with the same velocity. The impulse experienced by the body is

MV

1.5 MV

2 MV

zero

A Body of mass $= m$
velocity $= V (\hat{i})$
Knock out (Bounce velocity) $= V (- \hat{i})$
Change in momentum $= m (v) - (m) (- v)$
$= 2 M V$
Rate of change of momentum is Force
Thus $\frac{d p}{d t} =$ force.
force $\times$ time $=$ Impulse
Thus Impulse Experience by body $= 2 m v .$

Q. A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is

MV

1.5 MV

2 MV

zero

A Body of mass =m velocity =V(i^) Knock out (Bounce velocity) =V(−i^) Change in momentum =m(v)−(m)(−v) =2MV Rate of change of momentum is Force Thus dtdp​​= force. force × time = Impulse Thus Impulse Experience by body =2mv.

Q. A body of mass $M$ hits normally a rigid wall with velocity $V$ and bounces back with the same velocity. The impulse experienced by the body is

A Body of mass $= m$
velocity $= V (\hat{i})$
Knock out (Bounce velocity) $= V (- \hat{i})$
Change in momentum $= m (v) - (m) (- v)$
$= 2 M V$
Rate of change of momentum is Force
Thus $\frac{d p}{d t} =$ force.
force $\times$ time $=$ Impulse
Thus Impulse Experience by body $= 2 m v .$