Laws of Motion - NEET Physics Chapterwise MCQs & PYQs
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NEET Laws of Motion MCQs & PYQs
Practice NEET Laws of Motion Questions
Question 1:
easy
A particle of mass \( m \) is projected with velocity \( v \) making an angle of \( 45^{\circ} \) with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be:
(2008)
Horizontal momentum remains unchanged. Initial vertical momentum \( P_{iy} = m v \sin 45^{\circ} = m v / \sqrt{2} \). Final vertical momentum \( P_{fy} = -m v \sin 45^{\circ} = -m v / \sqrt{2} \). Change in vertical momentum \( \Delta P_y = P_{fy} - P_{iy} = -\sqrt{2} mv \). The magnitude of the change in momentum is \( \sqrt{2} mv \).
A \( 0.5 \text{ kg} \) ball moving with a speed of \( 12 \text{ m/s} \) strikes a hard wall at an angle of \( 30^{\circ} \) with the wall. It is reflected with the same speed at the same angle. If the ball is in contact with the wall for \( 0.25 \) seconds, the average force acting on the wall is (2006)
Change in momentum \( \Delta P = 2mv \sin\theta \) (perpendicular to wall). \( \Delta P = 2 \times 0.5 \text{ kg} \times 12 \text{ m/s} \times \sin 30^{\circ} = 6 \text{ Ns} \). Average force \( F = \Delta P / \Delta t = 6 \text{ Ns} / 0.25 \text{ s} = 24 \text{ N} \).
A \( 1 \text{ kg} \) stationary bomb is exploded in three parts having mass \( 1 : 1 : 3 \) respectively. Parts having same mass move in perpendicular direction with velocity \( 30 \text{ m s}^{-1} \), then the velocity of bigger part will be:
(2001)
Masses \( m_1 = 0.2 \text{ kg}, m_2 = 0.2 \text{ kg}, m_3 = 0.6 \text{ kg} \). Momentum of first two parts: \( P_1 = 0.2 \times 30 = 6 \text{ Ns} \), \( P_2 = 0.2 \times 30 = 6 \text{ Ns} \). Since they are perpendicular, their resultant momentum \( P_{12} = \sqrt{6^2 + 6^2} = 6\sqrt{2} \text{ Ns} \). By conservation of momentum, \( P_3 = P_{12} = 6\sqrt{2} \text{ Ns} \). Velocity of bigger part \( v_3 = P_3 / m_3 = (6\sqrt{2}) / 0.6 = 10\sqrt{2} \text{ m/s} \).
A particle is projected with velocity \( u \) makes an angle \( \theta \) w.r.t. horizontal. Now it breaks in two identical parts at highest point of trajectory. If one part is retrace its path, then velocity of other part is:
(1999)
At the highest point, velocity is \( u \cos\theta \) (horizontal) and mass is \( M \). Initial momentum \( P_i = M u \cos\theta \). Particle breaks into two identical parts (\( M/2 \) each). One part retraces its path, so its velocity is \( -u \cos\theta \). By conservation of momentum: \( M u \cos\theta = (M/2) (-u \cos\theta) + (M/2) v_2 \). Solving for \( v_2 \) gives \( v_2 = 3u \cos\theta \).
For a rocket propulsion velocity of exhaust gases relative to rocket is \( 2 \text{ km/s} \). If mass of rocket system is \( 1000 \text{ kg} \), then the rate of fuel consumption for a rocket to rise up with acceleration \( 4.9 \text{ m/s}^2 \) will be:
(1998)
Thrust force \( F_t = v_e \frac{dM}{dt} \). For upward motion with acceleration \( a \): \( F_t - Mg = Ma \). \( v_e \frac{dM}{dt} = M(g + a) \). \( \frac{dM}{dt} = \frac{M(g + a)}{v_e} = \frac{1000 \text{ kg} (9.8 + 4.9) \text{ m/s}^2}{2000 \text{ m/s}} = 7.35 \text{ kg/s} \).
If force \( F = 500 – 100t \), then function of impulse with time will be:
(1998)
Impulse \( J \) is the integral of force \( F \) with respect to time \( t \). \( J = \int F dt = \int (500 - 100t) dt \). \( J = 500t - 100 \frac{t^2}{2} + C \). Assuming \( J=0 \) at \( t=0 \), then \( C=0 \). So, \( J = 500t - 50t^2 \).
Which one of the following statements is incorrect?
(2018)
Friction force always opposes relative motion. Limiting static friction \( \text{f}_{\text{s,max}} \) is proportional to normal reaction \( \text{N} \), so \( \text{f}_{\text{s,max}} = \mu_{\text{s}}\text{N} \). Rolling friction is generally much smaller than sliding friction. The coefficient of friction \( \mu \) is a ratio of forces (friction force to normal force), thus it is a dimensionless quantity. Therefore, option (d) is incorrect.
A conveyor belt is moving at a constant speed of \( 2 \text{ m/s} \). A box is gently dropped on it. The coefficient of friction between them is \( \mu = 0.5 \). The distance that the box will move relative to belt before coming to rest on it, taking \( g = 10 \text{ ms}^{-2} \).
(2011 Mains)
Initial relative speed of box \( u = 2 \text{ m/s} \). Friction force \( f = \mu mg \). Acceleration \( a = \frac{f}{m} = \mu g = 0.5 \times 10 = 5 \text{ m/s}^2 \). Using \( v^2 = u^2 - 2as \) (since friction causes deceleration). With \( v = 0 \), distance \( s = \frac{u^2}{2a} = \frac{(2)^2}{2 \times 5} = \frac{4}{10} = 0.4 \text{ m} \).
A block of mass \(10\text{ kg}\) placed on rough horizontal surface having coefficient of friction \(\mu = 0.5\), if a horizontal force of \(100text{ N}\) acting on it then acceleration of the block will be:
(2002)
Calculate max friction \(f = \mu mg\). Compare with applied force (F). If (F > f), then (a = (F-f)/m). Here, (f = 0.5 \times 10 \times 10 = 50\text{ N}). So, (a = (100-50)/10 = 5\text{ m/s}^2).
On the horizontal surface of a truck a block of mass \(1\text{ kg}\) is placed \((\mu = 0.6)\) and truck is moving with acceleration \(5\text{ m/s}^2\) then the frictional force on block will be:
(2001)
Required force for block to accelerate with truck is \(F = ma_{truck}\). Maximum static friction is \(f_{s,max} = \mu mg\). If \(F \leq f_{s,max}\), friction equals (F). Here, \(F = 1 \times 5 = 5\text{ N}\) and \(f_{s,max} = 0.6 \times 1 \times 10 = 6\text{ N}\). Since \(5 \leq 6\), the frictional force is \(5\text{ N}\).