NEET 2016 Phase 2 | Center of Mass and Collision Question 14 | Physics

NEET 2016 Phase 2

MCQ (Single Correct Answer)

-1

A bullet of mass 10 g moving horizontally with a velocity of 400 m s^$$-$$1 strikes a wood block of mass 2 kg which is suspended by light inextensible string of length 5 m. As a result, the centre of gravity of the block found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be

100 m s^$$-$$1

80 m s^$$-$$1

120 m s^$$-$$1

160 m s^$$-$$1

NEET 2016 Phase 2

MCQ (Single Correct Answer)

-1

Two identical balls A and B having velocities of 0.5 m s^$$-$$1 and $$-$$0.3 m s^$$-$$1 respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be :

$$-$$0.5 m s^$$-$$1 and 0.3 m s^$$-$$1

0.5 m s^$$-$$1 and $$-$$0.3 m s^$$-$$1

$$-$$0.3 m s^$$-$$1 and 0.5 m s^$$-$$1

0.3 m s^$$-$$1 and 0.5 m s^$$-$$1

NEET 2016 Phase 2

MCQ (Single Correct Answer)

-1

A rigid ball of mass m strikes a rigid wall at 60^o and gets reflected without loss of speed as shown in the figure. The value of impulse imparted by the wall on the ball will be :
NEET 2016 Phase 2 Physics - Center of Mass and Collision Question 33 English

NEET 2016 Phase 2 Physics - Center of Mass and Collision Question 33 English

$$mV$$

2 $$mV$$

$${{mV} \over 2}$$

$${{mV} \over 3}$$

AIPMT 2015 Cancelled Paper

MCQ (Single Correct Answer)

-1

Two particles of masses m₁, m₂ move with initial velocities u₁ and u₂. On collision, one of the particles get excited to higher level, after absorbing energy $$\varepsilon $$. If final velocities of particles be v₁ and v₂ then we must have :

$${1 \over 2}$$m₁u$$_1^2$$ + $${1 \over 2}$$ m₂u$$_2^2$$ $$-$$ $$\varepsilon $$ = $${1 \over 2}$$ m₁v$$_1^2$$ + $${1 \over 2}$$m₂v$$_2^2$$

$${1 \over 2}$$m$$_1^2$$u$$_1^2$$ + $${1 \over 2}$$m$$_2^2$$u$$_2^2$$ + $$\varepsilon $$ = $${1 \over 2}$$m$$_1^2$$v$$_1^2$$ + $${1 \over 2}$$m$$_2^2$$v$$_2^2$$

m$$_1^2$$u₁ + m$$_2^2$$u₂ $$-$$ $$\varepsilon $$ = m$$_1^2$$v₁ + m$$_2^2$$v₂

$${1 \over 2}$$m₁u$$_1^2$$ + $${1 \over 2}$$m₂u$$_2^2$$ = $${1 \over 2}$$m₁v$$_1^2$$ + $${1 \over 2}$$m₂v$$_2^2$$ $$-$$ $$\varepsilon $$

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