Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a friction less horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is

30 m/s
20 m/s
10 m/s
5 m/s

Solution

\(V_{c}=\frac{10\times 14+4\times 0}{10+4}\)= 10m/s ; since spring force is internal force.

A particle of mass m₁ moving with velocity v strikes with amass m₂ at rest, then the condition for maximum transfer of kinetic energy is

m₁ >> m₂
m₂ >> m₂
m₁ = m₂
m₁ = 2m₂

Solution

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to

x
e^x
x²
loge x

Solution

Two masses m_a and m_b moving with velocities v_a and v_b in opposite direction collide elastically and after the collision m_a and m_b move with velocities V_b and V_a respectively.Then the ratio m_a/m_b is

\(\frac{V_{a}-V_{b}}{V_{a}+V_{b}}\)
\(\frac{m_{a}-m_{b}}{m_{a}}\)
1
¹⁄₂

Solution

As velocities are exchanged on perfectly elastic collision, therefore masses of two objects must be equal.

∴ ^m_a⁄_{m_b} = 1 or m_a = m_b

The bob A of a simple pendulum is released when the string makes an angle of 45º with the vertical. It hits another bob B of the same material and same mass kept at rest on the table. If the collision is elastic

both A and B rise to the same height
both A and B come to rest at B
both A and B move with the velocity of A
A comes to rest and B moves with the velocity of A

Solution

As bob B is of same material and same mass as the bob A, therefore, on elastic collision, their velocities are exchanged. Bob A comes to rest and B moves with the velocity of A.

A shell is fired from a cannon with a velocity V at an anglewith the horizontal direction. At the highest point in its path, it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon. The speed of the other piece immediately after the explosion is

3 V cos θ
2 V cos θ
³⁄₂V cos θ
V cos θ

Solution

A nucleus moving with a velocity \(\overrightarrow{v}\) emits an α-particle.Let the velocities of the α-particle and the remaining nucleus be \(\overrightarrow{v}_{1}\) and \(\overrightarrow{v}_{2}\) and their masses be m₁ and m₂.

\(\overrightarrow{v}\), \(\overrightarrow{v}_{1}\) and \(\overrightarrow{v}_{2}\) must be parallel to each other
None of the two of \(\overrightarrow{v}\), \(\overrightarrow{v}_{1}\) and \(\overrightarrow{v}_{2}\) should be parallel to each other
\(m_{1}\overrightarrow{v}_{1}+m_{2}\overrightarrow{v}_{2}\) must be parallel to (m₁+m₂)\(\overrightarrow{v}\)
None of these

Solution

A small block of mass m is kept on a rough inclined surface of inclination θ fixed in an elevator. The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t as seen by the observer on the inclined plane will be

zero
mgvt cos²θ
mgvt sin²θ
mgvt sin 2θ

Solution

Since block does not slide on wedge so displacement is zero & hence work done by force is zero.

A particle of mass m moving eastward with a speed v collides with an other particle of the same mass moving northwards with the same speed. If two particles coalesce on collision, the new particle of mass 2 m will move in the north-east direction with a velocity

v / 2
v√2
v/√2
None of these

Solution

A particle, initially at rest on a friction less horizontal surface,is acted upon by a horizontal force which is constant in magnitude and direction. A graph is plotted of the work done on the particle W,against the speed of the particle v.If there are no other horizontal forces acting on the particle,the graph would look like

Solution

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