Calculate the K.E and P.E. of the ball half way up, when a ball of mass 0.1 kg is thrown vertically upwards with an initial speed of 20ms^–1.

10 J, 20 J
10 J, 10 J
15 J, 8 J
8 J, 16 J

Solution

Total energy at the time of projection

= ¹⁄₂ m v² = ¹⁄₂ × 0.1 (20)² = 20j

Half wayup, P.E. becomes half the P.E. at the top i.e.

P.E.= ²⁰⁄₂= 10j ∴ K.E. = 20 –10 = 10J.

A small body is projected in a direction inclined at 45º to the horizontal with kinetic energy K. At the top of its flight, its kinetic energy will be

Zero
K/2
K/4
K/√2

Solution

Two bodies A and B having masses in the ratio of 3 : 1 possess the same kinetic energy.The ratio of linear momentum of B to A is

1 : 3
3 : 1
1 : √3
√3 : 1

Solution

A block is acted upon by a force, which is inversely proportional to the distance covered (x). The work done will be proportional to

x
x^1/2
x²
None of these

Solution

W = F × s

W ∝ ¹⁄_x (x) ∴ W ∝ x⁰

A particle moves under the effect of a force F = cx from x = 0 to x = x₁, the work done in the process is

\(cx_{1}^{2}\)
\(\frac{1}{2}cx_{1}^{2}\)
\(2cx_{1}^{2}\)
zero

Solution

A simple pendulum 1 metre long has a bob of 10 kg. If the pendulum swings from a horizontal position, the K.E. of the bob, at the instant it passes through the lowest position of its path is

89 joule
95 joule
98 joule
85 joule

Solution

A body of mass m accelerates uniformly from rest to v₁ in time t₁. As a function of t, the instantaneous power delivered to the body is

\(\frac{mv_{1}t}{t_{2}}\)
\(\frac{mv_{1}^{2}t}{t_{1}}\)
\(\frac{mv_{1}t^{2}}{t_{1}}\)
\(\frac{mv_{1}^{2}t}{t_{1}^{2}}\)

Solution

A metallic wire of length L metre extends by ι metre when stretched by suspending a weight Mg from it. The mechanical energy stored in the wire is

2 Mg ι
Mg ι
^Mgι⁄₂
^Mgι⁄₄

Solution

Weight Mg moves the centre of gravity of the spring

through a distance \(\frac{(0+\iota )}{2}=\iota /2\)

Mechanical energy stored = Work done = Mg ι/2

A cord is used to lower vertically a block of mass M,a distance d at a constant downward acceleration of g/4.The work done by the cordon the block is

Mg^d⁄₄
3 Mg^d⁄₄
-3 Mg^d⁄₄
Mg d

Solution

As the cord is trying to hold the motion of the block,work done by the cord is negative.

The vessels A and B of equal volume and weight are immersed in water to depth h. The vessel A has an opening at the bottom through which water can enter. If the work done in immersing A and B are W_A and W_B respectively,then

W_A = W_B
W_A < W_B
W_A > W_B
W_A ≠ W_B

Solution

Because water enters into the vessel A, it becomes heavier. Gravity helps it sink. External work required for immersing A is obviously less than that for immersing B.

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