Laws of Motion

Question 505
CBSEENPH11020440

A mass ‘m’ moves with a velocity v and collides inelastically with another identical mass. After collision, the 1st mass moves with velocity v/√3 in a direction perpendicular to the initial direction of motion. Find the speed of the 2nd mass after collision

  • v

  • √3 v

  • 2v/√3

  • v/√3

Solution

C.

2v/√3

mv= mv1 cos = θ

0 space equals space fraction numerator mv over denominator square root of 3 end fraction space minus mv subscript 1 sinθ
therefore space straight v subscript 1 space equals space fraction numerator 2 over denominator square root of 3 end fraction straight v

Question 506
CBSEENPH11020445

Consider a car moving on a straight road with a speed of 100 m/s. The distance at which car can be stopped is [µ =k 0.5]

  • 800 m

  • 1000 m

  • 600 m

  • 900 m

Solution

B.

1000 m

straight mu subscript straight k mgs space equals space 1 half mu squared
straight s equals space fraction numerator straight u squared over denominator 2 straight mu subscript straight k straight g end fraction space equals space 1000 space straight m
Question 507
CBSEENPH11020463

A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds?

  • h/9 metres from the ground

  • 7h/9 metres from the ground

  • 8h/9 metres from the ground

  • 17h/18 metres from the ground.

Solution

C.

8h/9 metres from the ground


second law of motion
straight s space equals ut space plus space 1 half space plus gT squared
or space straight h equals 0 plus 1 half gT squared space left parenthesis because space straight u equals 0 right parenthesis
therefore space straight T equals square root of open parentheses fraction numerator 2 straight h over denominator straight g end fraction close parentheses end root
At space straight t space equals space straight T over 3 straight s comma
straight s space equals space 0 plus space 1 half straight g space open parentheses straight T over 3 close parentheses squared
rightwards double arrow space straight s space equals space 1 half space straight g. straight T squared over 9
rightwards double arrow space straight s space equals space straight g over 18 space straight x space fraction numerator 2 straight h over denominator straight g end fraction space space space space open parentheses therefore space equals space square root of fraction numerator 2 straight h over denominator straight g end fraction end root close parentheses
therefore s = h/9 m
Hence, the position of ball from the ground= h- h/9 = 8h/9 m
Question 508
CBSEENPH11020467

An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 cm. If the car is going twice as fast, i.e 120 km/h, the stopping distance will be

  • 20 m

  • 40 m

  • 60 m

  • 80 m

Solution

D.

80 m

Third equation of motion gives
v2 = u2 + 2as ⇒ 2
s ∝ u (since v = 0)
where a = retardation of body in both the cases
therefore, straight s subscript 1 over straight s subscript 2 space equals space fraction numerator straight u subscript 1 superscript 2 over denominator straight u subscript 2 superscript 2 end fraction .... (i)
Here, s1 = 20 m, u1 = 60 km/h, u2 = 120 km/h. Putting the given values in eq. (i), we get
20 over straight s subscript 2 space equals space open parentheses 60 over 120 close parentheses squared
space straight s subscript 2 space equals space 20 space straight x space open parentheses 120 over 60 close parentheses squared
space equals space 20 space straight x space 4 space
space equals space 80 space straight m

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