Explanation
Hint:
· This question is directly formula based, in the question, radius of curvature is given, so here, in this case we need to use the formula of velocity directly.
· The equation of velocity is $${\rm{v = }}\sqrt {{\rm{gRtan}}\theta }$$
Step 1:Resolve the forces.
Width of the road is given,$${\rm{L = 10cm}}$$ Radius or curvature, $${\rm{R = 50cm}}$$ The distance between the above and lower edge of the road, $${\rm{h = 1}}{\rm{.5cm}}$$
By applying newtons second law of motion,
$$\Sigma F_{net}= m{a_c}$$
Here, $${a_c}$$ is the centripetal acceleration.
The components of the force will be:
$${\rm{Nsin}}\theta {\rm{ = }}\dfrac{{m{v^2}}}{{\rm{R}}}$$ ….....(1)
$${\rm{Ncos}}\theta {\rm{ = mg}}$$ ……(2)
Now, when we divide equation (1) by (2), we get,
$${\rm{tan}}\theta {\rm{ = }}\dfrac{{{v^2}}}{{gR}}$$
Step 2: Apply the formula of velocity
$${\rm{v = }}\sqrt {{\rm{gRtan}}\theta }$$ …..(3)
Also, we know that $${\rm{tan}}\theta {\rm{ = }}\dfrac{{\rm{h}}}{{\rm{L}}}$$
Now, we put the values in equation (3),
$$\Rightarrow {\rm{v = }}\sqrt {10 \times 50 \times \dfrac{{1.5}}{{10}}}$$
$$ \Rightarrow {\rm{v = }}\sqrt {50 \times 1.5}$$
$$ \Rightarrow {\rm{v = }}\sqrt {75} $$
$$ \Rightarrow {\rm{v = 8}}{\rm{.5m/sec}}$$
Hence, option (D), $${\rm{8}}{\rm{.5m/sec}}$$is the correct answer.
THIS QUESTION IS NOT CLEARED.
Given that,
Mass =$$m$$
Initial velocity =$$v_ {1} $$
Final velocity=$$v_ {2} $$
We know that,
Impulse is the change in momentum.
So, the impulse is
$$I=\Delta p$$.....(I)
$$\Delta p=m\Delta v$$
Put the value of $$\Delta p$$ in equation (I)
$$I =m\Delta v$$
$$\Delta v$$ is the change in velocity
The impulse will be
$$I=m (v_ {2}-v_ {1}) $$
The impulse will be $$I= m (v_ {2}-v_ {1}) $$
Hence, this is the required solution.
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