Work, Energy and Power
- A metal ball of mass 2 kg moving with a velocity of 36 km/h has a head on collision with a stationary ball of mass 3 kg. If after the collision, the two balls move together, the loss in kinetic energy due to collision is
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Applying conservation of momentum, m1v1 = (m1 + m2)v
= m1v1 (m1 + m2)
Here, v1 = 36 km/hr = 10 m/s
m1 = 2 kg, m2 = 3 kgv = 10 × 2 = 4 m/s 5 K.E. (initial) = 1 × 2 × ( 10 )2 = 100 J 2 K.E. (Final) = 1 × (3 + 2) × ( 4 )2 = 40 J 2
Loss in K.E. = 100 – 40 = 60 J
Correct Option: C
Applying conservation of momentum, m1v1 = (m1 + m2)v
= m1v1 (m1 + m2)
Here, v1 = 36 km/hr = 10 m/s
m1 = 2 kg, m2 = 3 kgv = 10 × 2 = 4 m/s 5 K.E. (initial) = 1 × 2 × ( 10 )2 = 100 J 2 K.E. (Final) = 1 × (3 + 2) × ( 4 )2 = 40 J 2
Loss in K.E. = 100 – 40 = 60 J
- A molecule of mass m of an ideal gas collides with the wall of a vessel with a velocity v and returns back with the same velocity. The change in linear momentum of molecule is
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Correct Option: A
- Two equal masses m1 and m2 moving along the same straight line with velocities + 3 m/s and – 5m/s respectively, collide elastically. Their velocities after the collision will be respectively.
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In elastic collision, the velocities get inter changed if the colliding objects have equal masses.
Correct Option: D
In elastic collision, the velocities get inter changed if the colliding objects have equal masses.
- A bomb of mass 1 kg is thrown vertically upwards with a speed of 100 m/s. After 5 seconds it explodes into two fragments. One fragment of mass 400 gm is found to go down with a speed of 25 m/s. What will happen to the second fragment just after the explosion? (g = 10 m/s2)
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Speed of bomb after 5 second, v = u – gt = 100 –10 × 5 = 50m/s
Momentum of 400 g fragment = 400 × (-25) [downward] 1000 Momentum of 600g fragment = 600 v 1000
Momentum of bomb = 1 × 50 = 50
From conservation of momentum
Total momentum before splitting = total momentum after splitting.⇒ 50 = - 400 × 25 + 600 v 1000 1000
⇒ v = 100 m/s [upward]Correct Option: B
Speed of bomb after 5 second, v = u – gt = 100 –10 × 5 = 50m/s
Momentum of 400 g fragment = 400 × (-25) [downward] 1000 Momentum of 600g fragment = 600 v 1000
Momentum of bomb = 1 × 50 = 50
From conservation of momentum
Total momentum before splitting = total momentum after splitting.⇒ 50 = - 400 × 25 + 600 v 1000 1000
⇒ v = 100 m/s [upward]
- A stationary particle explodes into two particles of masses m1 and m2 which move in opposite directions with velocities v1 and v2. The ratio of their kinetic energies E1 / E2 is
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From conservation law of momentum, before collision and after collision linear momentum (p) will be same. That is,
initial momentum = final momentum.
⇒ 0 = m1v1 – m2v2 ⇒ m1v1 = m2v2
p1 = p2E = p2 2m ∴ E1 = p12 × 2m2 E2 2m1 p22 ⇒ E1 = m2 [p1 = p2] E2 m1 Correct Option: B
From conservation law of momentum, before collision and after collision linear momentum (p) will be same. That is,
initial momentum = final momentum.
⇒ 0 = m1v1 – m2v2 ⇒ m1v1 = m2v2
p1 = p2E = p2 2m ∴ E1 = p12 × 2m2 E2 2m1 p22 ⇒ E1 = m2 [p1 = p2] E2 m1
