Skip to content
Question of 77

Q.(a) What is Inertia? Write its types and discuss one example of any one type of Inertia.

(b) A bullet of 50g is fired from a gun of 2kg with a velocity of 40 m/s. Find the recoil velocity of the gun.
(OR)
(a) What is banking of roads? Derive an expression for maximum speed of vehicle on the banked road. [1+3 marks]
(b) The wheels of automobiles are made circular. Why? [1 mark]
Himachal HpboseHPBOSE Himachal Pradesh Class 11 Board Exam 2025Subjective· 5mImportance★★★★★
0% · 0/77 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

(a) Inertia has 3 types — rest, motion, direction. (b) By conservation of momentum, the gun recoils at 1 m/s.

(a) Inertia

Inertia is the natural property of a body by virtue of which it resists any change in its state of rest, or of uniform motion in a straight line, unless it is acted upon by an external unbalanced force. Inertia is a measure of a body's reluctance to change its state of motion, and it depends on the mass of the body (larger mass → larger inertia).

There are three types of inertia:

  1. Inertia of rest — the tendency of a body to continue in its state of rest.
  2. Inertia of motion — the tendency of a body to continue in its state of uniform motion.
  3. Inertia of direction — the tendency of a body to continue moving in the same direction (resisting any change of direction).

Example (inertia of rest): When a bus at rest suddenly starts moving, a standing passenger tends to fall backward. This happens because the lower part of the passenger's body (in contact with the bus floor) is set into motion along with the bus, but the upper part of the body tends to remain at rest due to inertia of rest — so relative to the bus, the passenger appears to be thrown backward.

(b) Numerical — recoil velocity of the gun

Given: mass of bullet m1=50 g=0.05 kgm_1 = 50\ \text{g} = 0.05\ \text{kg}, mass of gun m2=2 kgm_2 = 2\ \text{kg}, velocity of bullet after firing v1=40 m/sv_1 = 40\ \text{m/s}.

Before firing, the gun and bullet are both at rest, so total initial momentum = 0.

By conservation of linear momentum (no external horizontal force acts on the gun-bullet system during firing):

m1v1+m2v2=0m_1 v_1 + m_2 v_2 = 0

(0.05)(40)+(2)v2=0(0.05)(40) + (2)v_2 = 0

2+2v2=02 + 2v_2 = 0

v2=−1 m/sv_2 = -1\ \text{m/s}

The negative sign shows the gun recoils in the direction opposite to the bullet's motion. So the recoil speed of the gun is 1 m/s.

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.