Question 205: Laws of Motion

Question

A block of mass $$m$$ is moving on rough horizontal surface with momentum $$p$$. The coefficient of friction between the block and surface is $$\mu$$. The distance covered by block before it stops is [$$g=$$ acceleration due to gravity]

Accepted Answer

Using kinematic relation,
$$v^2=u^2-2 a s \quad 	ext{.... (i)}$$
The momentum $$p$$ is given by
$$\begin{aligned}
p & =M u \
u & =\frac{p}{M}
\end{aligned}$$
and acceleration is given as $$a=\mu g$$
Substituting values in Eq. (i), we get
$$0=u^2-2 	ext { as } \quad(\because 	ext { final velocity }, v=0)$$
$$0=\left(\frac{p}{M}ight)^2-2 \mu g s$$
$$\left(\frac{p}{M}ight)^2=2 \mu g s$$
$$\Rightarrow \quad s=\frac{p^2}{2 M^2 \mu g}$$

Answer

$$\frac{2 \mu M^2 g}{p^2}$$

Answer

$$\frac{p}{2 \mu M g}$$

Answer

$$\frac{2 \mu M g}{p}$$

Answer

$$\frac{p^2}{2 \mu M^2 g}$$

Laws of Motion

A block of mass $m is moving on rough horizontal surface with momentum p. The coefficient of friction between the block and surface is \mu. The distance covered by block before it stops is [g=$ acceleration due to gravity]

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