A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m = 0 . 4 ?kg is at rest on this surface. An impulse of 1 . 0 Ns is applied to the block at time t = 0 so that it starts moving along the x -axis with a ve

Question

A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass $m = 0.4 kg$ is at rest on this surface. An impulse of $1.0 Ns$ is applied to the block at time $t = 0$ so that it starts moving along the $x$ -axis with a velocity $v (t) = v_{0} e^{- t / τ}$ , where $v_{0}$ is a constant and $τ = 4 s$ . The displacement of the block, in metres, at $t = τ$ is _____. Take $e^{- 1} = 0.37$ . [2017]

Smart Achievers · Accepted Answer

(6.30)

$Given: mass m = 0.4 kg; impulse J = 1.0 Ns$ $V_{(t)} = v_{0} e^{- t / τ}; τ = 4 s and e^{- 1} = 0.37$

$Impulse = Change in linear momentum$

$or, J = m V_{0} \Rightarrow V_{0} = \frac{J}{m} = \frac{1}{0.4} = 2.5 {ms}^{- 1}$

$Also (V_{t}) = v_{0} e^{- t / τ}$ $∴$ $\frac{d s}{d t} = v_{0} e^{- t / τ} \Rightarrow d s = v_{0} e^{- t / τ} d t$

$∴$ $s = v_{0} \int_{0}^{τ} e^{- t / τ} d t = v_{0} τ (1 - e^{- 1}) = 2.5 \times 4 \times 0.63 = 6.30 m$

Solution

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Solution

Ans. (6.30) Given: mass m=0.4 kg; impulse J=1.0 Ns V(t)=v0e-t/τ; τ=4 s and e-1=0.37 Impulse = Change in linear momentum or, J=mV0⇒V0=Jm=10.4=2.5 ms-1 Also (Vt)=v0e-t/τ ∴ dsdt=v0e-t/τ⇒ds=v0e-t/τ dt ∴ s=v0∫0τe-t/τ dt=v0τ(1-e-1)=2.5×4×0.63=6.30 m

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