A cyclist moving at a speed of 20 m/s takes a turn, if he doubles his speed then chance of overturn

is doubled
is halved
becomes four times
becomes 1/4 times

Solution

When a cyclist moves on a circular path, it experiences a centrifugal force which is equal to mv²/ r.It tries to overturn the cyclist in outward direction. If speed increases twice, the value of centrifugal force too increases to 4 times its earlier value.Therefore the chance of overturning is 1/4 times.

A projectile is thrown in the upward direction making an angle of 60° with the horizontal direction with a velocity of 147 ms^-1. Then the time after which its inclination with the horizontal is 45°, is

15 s
10.98 s
5.49 s
2.745 s

Solution

A particle moves in a circle of radius 4 cm clockwise at constant speed 2 cm/s. If \(\hat{x}\) and \(\hat{y}\) are unit acceleration vectors along X and Y-axis respectively (in cm/s²), the acceleration of the particle at the instant half way between P and Q is given by

\(-4(\hat{x}+\hat{y})\)
\(4(\hat{x}+\hat{y})\)
\(-(\hat{x}+\hat{y})/\sqrt{2}\)
\((\hat{x}-\hat{y})/4\)

Solution

a=^V²⁄_r= 1 cm/s. Centripetal acceleration is directed towards the centre. Its magnitude = 1. Unit vector at the mid point on the path between P and Q is \(-(\hat{x}+\hat{y})/\sqrt{2}\).

A person sitting in the rear end of the compartment throws a ball towards the front end. The ball follows a parabolic path. The train is moving with velocity of 20 m/s. A person standing outside on the ground also observes the ball. How will the maximum heights (y_m) attained and the ranges (R)seen by the thrower and the outside observer compare with each other?

Same y_m different RSame y_m different R
Same y_m and R
Different y_m same R
Different y_m and R

Solution

The motion of the train will affect only the horizontal component of the velocity of the ball. Since, vertical component is same for both observers, the y_m will be same, but R will be different

A projectile of mass m is thrown with a velocity v making an angle 60° with the horizontal. Neglecting air resistance, the change in momentum from the departure A to its arrival at B,along the vertical direction is

2 mv
√3 mv
mv
^mv⁄_√3

Solution

As the figure drawn above shows that at points A and B the vertical component of velocity is v sin 60° but their directions are opposite.Hence, change in momentum is given by :

Δp=mv sin60°-(-mv sin60°)=2mv sin60°

=2mv^√3⁄₂=√3mv

A bullet is fired with a speed of 1500 m/s in order to hit a target 100 m away. If g = 10 m/s². The gun should be aimed

15 cm above the target
10 cm above the target
2.2 cm above the target
directly towards the target

Solution

The equation of trajectory of projectile is given by \(y=\frac{x}{\sqrt{3}}-\frac{gx^{2}}{20}\), where x and y are in metre.The maximum range of the projectile is

⁸⁄₃m
⁴⁄₃m
³⁄₄m
³⁄₈m

Solution

gun is aimed at a horizontal target. It takes 21s for the bullet to reach the target. The bullet hits the target x metre below the aim. Then, x is equal to

^9.8⁄₄m
^9.8⁄₈m
9.8 m
19.6 m

Solution

The velocity of projection of oblique projectile is \((6\hat{i}+8\hat{j})\) \((6\hat{i}+8\hat{j})\) ms^-1. The horizontal range of the projectile is

4.9 m
9.6 m
19.6 m
14 m

Solution

A projectile A is thrown at an angle of 30° to the horizontal from point P.At the same time, another projectile B is thrown with velocity v₂ upwards from the point Q vertically be low the highest point. For B to collide with A,\(\frac{V_{2}}{V_{1}}\) should be

1
2
¹⁄₂
4

Solution

This happen when vertical velocity of both are same.

∴ v₂=v₁ sin30° or \(\frac{V_{2}}{V_{1}}=\frac{1}{2}\)

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