Olympiad Test: Force and Laws of Motion - Class 9 MCQ

= 10 × (0.2 - 20 )

= 10 × (-19.8)

= -198 Ns.

5 = m × 20 ⇒ m = 0.25 kg

5 = (M + m) × a ⇒ 5 = (0.5 + 0.25) × a

⇒ a = 5/0.75 = 500/75 = 6.6 ms^-2

u₁ = 0, u₂ = 0 (both pistol and bullet are at rest)

v₁ = + 150 ms^–1 (the find velocity of the bullet)

let v₂ be the recoil velocity of the pistol

according to the law of conservation, total

moment remains the same.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

.02 × 0 + 2 × 0 = 0.02 × 150 + 2 × v²

0 = 3 + 2v²

⇒ v² = -3/2 = – 1.5 ms^–1

Negative sign indicates that the pistol will

recoil in the opposite direction of the bullet.

⇒ F = m₁a₁ ⇒ m₁ = 5/8, f = m₂a₂ ⇒ m₂ = 5/24

For two bodies combined, f = (m1 + m2) a

m1 + m2= 5/8 + 5/24 = 20/24 = 5/6 kg. f = 5N

∴ 5 = 5/6a ⇒ a = 6 ms^-2

u₁ = 0, u2 = 0 (both pistol and bullet are at rest)

v₁ = + 150 ms^–1 (the find velocity of the bullet)

let v2 be the recoil velocity of the pistol

according to the law of conservation, total

moment remains the same.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

.02 × 0 + 2 × 0 = 0.02 × 150 + 2 × v₂

0 = 3 + 2v₂

⇒ v₂ = -3/2 = – 1.5 ms^–1

Negative sign indicates that the pistol will recoil in the opposite direction of the bullet.

collision, they will move with a common velocity v given by

V =m_au_a + m_bu_b/m_a + m_b

Here, m_a = 20 kg, u_a = 2 m/s, m_b = 2 kg, u_b = 0

∴ v = 20 × 2 +2 × 0/20+2 = 40/22 = 1.81 m/s

impulse = force × time Ns
force=massx acc
impulse=massxacc.xtime [ acc.xtime= Change in velocity ]
impulse = massx change in velocity
             =0.15x(12-(-12))
             =0.15x24
             =3.6 Ns