(C) The correct relation is $\overrightarrow{F} \Delta t = m \Delta \overrightarrow{v}$,which represents the impulse-momentum theorem.From this,we can

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$\overrightarrow{F} \Delta t = m \Delta \overrightarrow{v}$

(C) The correct relation is $\overrightarrow{F} \Delta t = m \Delta \overrightarrow{v}$,which represents the impulse-momentum theorem.From this,we can write $F = \frac{m \Delta \overrightarrow{v}}{\Delta t}$.When you bend your knees upon landing,the time interval $\Delta t$ during which your velocity changes to zero increases.Since the change in momentum $m \Delta \overrightarrow{v}$ is constant,increasing the time $\Delta t$ results in a decrease in the impulsive force $F$ exerted on your knees,thereby reducing the risk of injury.

Class 11 Physics Newton's Laws of Motion and Friction Third Law of Motion and Momentum and Impulse

Easy

Consider the following statement: When jumping from some height,you should bend your knees as you come to rest,instead of keeping your legs stiff. Which of the following relations can be useful in explaining the statement?

A
$\Delta \overrightarrow{P_1} = - \Delta \overrightarrow{P_2}$
B
$\Delta E = - \Delta (PE + KE) = 0$
C
$\overrightarrow{F} \Delta t = m \Delta \overrightarrow{v}$
D
$\Delta \overrightarrow{x} \propto \Delta \overrightarrow{F}$

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Chapter

Newton's Laws of Motion and Friction

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Third Law of Motion and Momentum and Impulse

In an hour-glass,approximately $100$ grains of sand fall per second (starting from rest); and it takes $2 \, s$ for each sand particle to reach the bottom of the hour-glass. If the average mass of each sand particle is $0.2 \, g$,then the average force exerted by the falling sand on the bottom of the hour-glass is close to .......... $N$.

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$A$ body of mass $m_1$ exerts a force on another body of mass $m_2$. If the magnitude of acceleration of $m_2$ is $a_2$,then the magnitude of the acceleration of $m_1$ is (considering only two bodies in space).

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$A$ particle of mass $2 \ kg$ is initially at rest. $A$ force acts on it whose magnitude changes with time. The force-time graph is shown below. The velocity of the particle after $10 \ s$ is $.... \ ms^{-1}$.

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$A$ block of mass $2 \ kg$ is free to move along the $x$-axis. It is at rest and from $t=0$ onwards it is subjected to a time-dependent force $F(t)$ in the $x$ direction. The force $F(t)$ varies with $t$ as shown in the figure. The kinetic energy of the block after $4.5 \ s$ is (in $J$)

DifficultIIT 2010

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$A$ particle of mass $m \, kg$ moving with a velocity $v \, m/s$ strikes a surface as shown in the figure and rebounds with the same speed. The magnitude of the change in momentum is:

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Consider the following statement: When jumping from some height,you should bend your knees as you come to rest,instead of keeping your legs stiff. Which of the following relations can be useful in explaining the statement?

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$A$ body of mass $m_1$ exerts a force on another body of mass $m_2$. If the magnitude of acceleration of $m_2$ is $a_2$,then the magnitude of the acceleration of $m_1$ is (considering only two bodies in space).

$A$ particle of mass $2 \ kg$ is initially at rest. $A$ force acts on it whose magnitude changes with time. The force-time graph is shown below. The velocity of the particle after $10 \ s$ is $.... \ ms^{-1}$.

$A$ block of mass $2 \ kg$ is free to move along the $x$-axis. It is at rest and from $t=0$ onwards it is subjected to a time-dependent force $F(t)$ in the $x$ direction. The force $F(t)$ varies with $t$ as shown in the figure. The kinetic energy of the block after $4.5 \ s$ is (in $J$)

$A$ particle of mass $m \, kg$ moving with a velocity $v \, m/s$ strikes a surface as shown in the figure and rebounds with the same speed. The magnitude of the change in momentum is:

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