Question 267: Laws of Motion

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 $$v\left( t ight) = {v_0}{e^{ - t/	au }},$$ where $${v_0}$$ is a constant and $$	au  = 4s.$$ The displacement of the block, in
metres,
at $$t = 	au $$ is ______________ Take $${e^{ - 1}} = 0.37.$$

Accepted Answer

The block was initially at rest and its velocity just after the application of impulse is $$v(0) = {v_0}{e^{ - 0/	au }} = {v_0}$$. The applied impulse is equal to the change in linear momentum of the block i.e., J = mv
0
, which gives
v
0
= J/m = 1/0.4 = 2.5 m/s.
The velocity of the particle is given as
$$v(t) = {v_0}{e^{ - t/	au }}$$.
Integrate to get the displacement
$$x(t) = \int_0^t {{v_0}{e^{ - t/	au }}dt = {v_0}	au (} 1 - {e^{ - t/	au }})$$.
Substitute t = $$	au$$ = 4 s and v
0
= 2.5 m/s to get x($$	au$$) = (2.5) (4) (1 $$-$$ e
$$-$$1
) = 6.3 m.

Laws of Motion

Integer / Numeric Answer Type