Question

1. A light and a heavy object have the same momentum. Find out the ratio of
their kinetic energies. Which one has a larger kinetic energy?

2. An automobile engine propels a 1000 kg car (A) along a levelled road at a
speed of 36 km h–1. Find the power if the opposing frictional force is 100 N.
Now, suppose after travelling a distance of 200 m, this car collides with
another stationary car (B) of same mass and comes to rest. Let its engine
also stop at the same time. Now car (B) starts moving on the same level
road without getting its engine started. Find the speed of the car (B) just
after the collision.

3. A girl having mass of 35 kg sits on a trolley of mass 5 kg. The trolley is
given an initial velocity of 4 m s–1 by applying a force. The trolley comes to
rest after traversing a distance of 16 m. (a) How much work is done on the
trolley? (b) How much work is done by the girl?

4. Four men lift a 250 kg box to a height of 1 m and hold it without raising or
lowering it. (a) How much work is done by the men in lifting the box?

(b) How much work do they do in just holding it? (c) Why do they get tired

while holding it? (g = 10 m s–²)​

Answer 1

$\large\color{purple}\underline{\underline{{ \boxed{ Answer 01 }}}}$

 

$\tt \: Let \: `m_L \: and \: m_H` \: be \: the \: masses \: of \: the \: light \: and \: heavy \: object \\ \tt \: let \: `upsilon_L \: and \: upsilon_H` \: be \: their \: respective \: velocity \: \\ \\ \tt since \: momentum \: of \: light \: object = momentum \: of \: heavy \: object \\ \tt `m_L \: upsilon_L = m_H \: upsilon _H \\ \tt \: (m_H)/(m_L) (i)` \\ \tt \: Kinetic \: energy \: of \: light \: object, \\ \tt \: `(KE)_L = (1)/(2)m_L \: upsilon_L^2` \\ \tt \: Kinetic \: energy \: of \: heavy \: object, \\ \tt \: `(KE)_H = (1)/(2) m_H upsilon_H^2` \\ \\ \tt \: Clearly, `((KE)_L)/((KE)_H) = ((1)/(2)m_Lupsilon_L^2)/((1)/(2)m_H upsilon_H^2) \\ \tt \: = ((M_L)/(m_H)) ((upsilon_L)/(upsilon_H))^2 . \: . \: . \: . \: (ii)` \\ \tt \:frm \: eqn \: (i) and (ii) \\ \tt `((KE)_L)/((KE)_H) = (m_L)/(m_H) ((m_H)/(m_L))^2 \\ \tt= (m_H)/(m_L)` \\ \\ \tt \: Since `m_H gt m_L, (KE)_L gt (KE)_H` \\ \tt \: `((KE)_L)/((KE)_H) = (m_H)/(m_L) = (upsilon_L)/(upsilon_H)`$

$\large\color{purple}\underline{\underline{{ \boxed{ Answer 02 }}}}$

Given

• m(A)= m(B) =1000 kg.
• v = 36 km/h =10 m/s
• Frictional force = 100 N

Solution:

Since, the car A moves with a uniform speed, Therefore,

Before collision

$\tt \: Power = Force × \frac {distance} { time }\\ \tt \: = F . V = 100 N × 10 m/s \\ \tt \: = 1000 W$

After collision

$\tt \:{ \green{ m_A u_A + m_B u_B = m_A v_A + m_B v_B}} \\ \tt \: 1000 ×10 +1000 × 0 \\ \tt \: = 1000 × 0 + 1000 × v_B v_B \\ \tt \: = 10 m s–1$

$\large\color{purple}\underline{\underline{{ \boxed{ Answer 03 }}}}$

Given:

• Mass of girl = 35 kg
• Mass of trolley = 5 kg
• Total mass, m = (35 + 5) = 40 kg
•  u = 4 m/s
• v = 0
• s = 16 m

By equation,

$\tt \: v  {}^{2}  = u  {}^{2}  + 2as \\\tt \: 0 = (4)2 + 2a (16)\\\tt \:32a = –16\\\tt \:a = –0.5 m/s2$

Force exerted on trolley,

$\tt \: { \green{F = ma}} \\ \tt \: = 40 × 0.5 = 20 N$

 

Work done on trolley

$\tt \: W = FS \\ \tt \: = 20 N \times 16 m \\ \tt \: = 320 J$

Work done by the girl,

$\tt \: W = FS \\ \tt \: = mass \: of \: girl × retardation × S \\ \tt \:= 35 × 0.5 × 16 \\ \tt \: = 280 J$

$\large\color{purple}\underline{\underline{{ \boxed{ Answer 04 }}}}$

$\tt \: (a) F = 250 kg × g \\ \tt \:= 250 × 10 = 2500 N s \\ \tt \:= 1 m W \\ \\ \tt \:= F \times s = 2500 N m \\ \tt \:= 2500 J$

(b) Zero, as the box does not move at all while holding it.

(c) In order to hold the box men are applying a force which is opposite and equal to the gravitational force acting on the box. So they get tired.

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$\sf \: @WildCat7083$