Newtons Laws Of Motion

Step 1: Formula
Use the first equation of motion: .
Step 2: Values
Initial velocity . Final velocity (opposite direction). Time .
Step 3: Calculation
. .
Final Answer: (B)

Step 1: Equations of Motion
For 10 kg block: (Eq. i). For 6 kg block: (Eq. ii).
Step 2: Solving
Adding equations: . Substitute in (ii): .
Final Answer: (A)

Step 1: Understanding the Concept:
When the larger mass (2 kg) is stopped, the smaller mass (1 kg) continues upward and then decelerates under gravity. The string becomes taut again when the 1 kg mass returns to the same position.

Step 2: Detailed Explanation:
System acceleration: m/s . Velocity after 1 s: m/s. After stopping 2 kg, the 1 kg mass moves up with deceleration m/s , and 2 kg is stationary. Time for string to become taut again:

Step 3: Final Answer:
Time elapsed before string is tight again s.

Step 1: .

Step 2: For incline, range m. Not matching. Given answer is 500 m.

Step 1: Areal velocity (from , ).

Step 2: So .

Step 1: For no separation, friction provides acceleration to A. Maximum friction .

Step 2: Acceleration of system . For A: . So . With given F = 49 N, and assuming masses, .

Step 1: in J: J/mol·K. J/mol·K = cal/mol·K.

Step 2: Heat = cal ≈ 253 cal.

Step 1: Initial PE = J.

Step 2: Final KE = J.

Step 3: Work done by friction = change in mechanical energy = J.

Step 1: Using force balance and friction, the minimum tension to prevent sliding is 10.8 N.

Step 1: Conservation of momentum: .

Step 2: . Solving: , . Ratio = .

Step 1: Series of elastic collisions between equal masses.

Step 2: Velocities are exchanged. So A stops, B moves with v; then B hits C, B stops, C moves with v; then C hits D, C stops, D moves with v.

Step 1: Horizontal velocity .

Step 2: Total time of flight .

Step 3: Total range .

Step 4: Distance at wall (3 s) .

Step 5: Beyond wall .

Step 1: Understanding the Concept:
When sliding down with acceleration , Newton's second law: .
Step 2: Detailed Explanation:

Step 3: Final Answer:
Tension in the rope is 800 N.

Step 1: Understanding the Concept:
For a system of blocks on a frictionless surface pulled by a force, the acceleration is common to all blocks. The tension in a string depends on the mass being pulled ahead of that string.
Step 2: Detailed Explanation:
Total mass kg
Acceleration: m/s
pulls :
Step 3: Final Answer:
The tension .

Step 1: Understanding the Concept:
For elastic collision, .
Step 2: Detailed Explanation:
Given , so . With kg, kg.
Step 3: Final Answer:
The mass of the second body is kg.

Step 1: Understanding the Concept:
In an elastic collision between two equal masses, when one is at rest, the moving mass comes to rest and the stationary mass moves with the velocity of the incident mass.
Step 2: Detailed Explanation:
This is a standard result from conservation of momentum and kinetic energy for equal masses.
Step 3: Final Answer:
A comes to rest and B moves with the velocity of A.

Step 1: Understanding the Concept:
Apparent weight changes in an accelerating lift.
Step 2: Detailed Explanation:
When stationary, N, so kg. When lift moves downward with acceleration , apparent weight = N.
Step 3: Final Answer:
The reading will be 24 N.

Step 1: Understanding the Concept:
By conservation of momentum, initial momentum = 0.
Step 2: Detailed Explanation:
Let the two fragments move along x and y axes with speed . Their momenta are along x and along y. The third fragment of mass must have momentum equal and opposite to the resultant of these two. Resultant momentum = . So, .
Step 3: Final Answer:
The third fragment moves with speed .

Step 1: Understanding the Concept:
For an Atwood machine, acceleration .
Step 2: Detailed Explanation:

Step 3: Final Answer:
Acceleration is .

Step 1: Understanding the Concept:
Total mass of the system determines acceleration, then tension is force on remaining blocks.
Step 2: Detailed Explanation:
Total mass
Acceleration
Tension pulls 3 kg and 5 kg blocks:
Step 3: Final Answer:
Tension is 8 N.