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4-1.Newton's Laws of Motion
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A mass of $100\,g$ strikes the wall with speed $5\,m/s$ at an angle as shown in figure and it rebounds with the same speed. If the contact time is $2 \times {10^{ - 3}}\,\sec $, what is the force applied on the mass by the wall
A$250\sqrt 3 $N to right
B$250\, N$ to right
C$250\sqrt 3 $N to left
D$250 \,N$ to left
Solution
(c)Force = Rate of change of momentum
Initial momentum ${\vec P_1} = mv\sin \theta \;\hat i + mv\cos \theta \,\hat j$
Final momentum ${\vec P_2} = – mv\sin \theta \;\hat i + mv\cos \theta \,\hat j$
$\vec F = \frac{{\Delta \vec P}}{{\Delta t}} = \frac{{ – 2mv\sin \theta }}{{2 \times {{10}^{ – 3}}}}$
Substituting $m = 0.1 kg, v = 5 m/s$ , $\theta = 60°$
Force on the ball $\vec F = – 250\sqrt 3\, N$
Negative sign indicates direction of the force
Initial momentum ${\vec P_1} = mv\sin \theta \;\hat i + mv\cos \theta \,\hat j$
Final momentum ${\vec P_2} = – mv\sin \theta \;\hat i + mv\cos \theta \,\hat j$
$\vec F = \frac{{\Delta \vec P}}{{\Delta t}} = \frac{{ – 2mv\sin \theta }}{{2 \times {{10}^{ – 3}}}}$
Substituting $m = 0.1 kg, v = 5 m/s$ , $\theta = 60°$
Force on the ball $\vec F = – 250\sqrt 3\, N$
Negative sign indicates direction of the force
Standard 11
Physics
