A mass of \(0.5\) kg moving with a speed of \(1.5\) m/s on a horizontal smooth surface, collides with a nearly weightless spring with force constant \(k=50\) N/m. The maximum compression of the spring would be:
           
1.  \(0.12\) m
2.  \(1.5\) m
3.  \(0.5\) m
4.  \(0.15\) m

A block of mass m initially at rest, is dropped from a height h onto a spring of force constant k. If the maximum compression in the spring is x, then:     
         

1. $mgh=\frac{1}{2}k{x}^2$

2. $mg(h+x)=\frac{1}{2}k{x}^2$

3. $mgh=\frac{1}{2}k{(x+h)}^2$

4. $mg(h+x)=\frac{1}{2}k{(x+h)}^2$

A block of mass 2 kg moving with a velocity of 10 m/s on a smooth surface hits a spring of force constant $80\times {10}^3$ N/m as shown. The maximum compression in the spring will be:
               

A spring 40 mm long is stretched by the application of force. If 10 N force is required to stretch the spring through 1 mm, then work done to stretch the spring 40 mm is equal to:

When a spring is subjected to 4 N force, its length is a metre and if 5 N is applied, its length is b metre. If 9 N is applied, its length will be:

1.  4b – 3a

2.  5b – a

3.  5b – 4a

4.  5b – 2a

A block of mass \(\text M\) moving on the frictionless horizontal surface collides with the spring of spring constant \(\text K\)and compresses it by length \(\text L\). The maximum momentum of the block after the collision will be:
    

If two springs, A and B $(K_A$ $=$ $2$ $K_B),$ are stretched by the same suspended weights, then the ratio of work done in stretching is equal to:
1.  1 : 2
2.  2 : 1
3.  1 : 1
4.  1 : 4

A weight 'mg' is suspended from a spring. The energy stored in the spring is U. The elongation in the spring is:

1.  $\frac{2\mathrm{U}}{\mathrm{mg}}$

2.  $\frac{\mathrm{U}}{\mathrm{mg}}$

3.  $\frac{\sqrt{2}\mathrm{U}}{\mathrm{mg}}$

4.  $\frac{\mathrm{U}}{\sqrt{2}\mathrm{mg}}$

Two springs of spring constants k and 3k are stretched separately by the same force. The ratio of potential energy stored in them respectively, will be: 

A vertical spring with a force constant \(k\) is fixed on a table. A ball of mass \(m\) at a height \(h\) above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance \(d\). The net work done in the process is:
1. \(mg(h+d)+\frac{1}{2}kd^2\)
2. \(mg(h+d)-\frac{1}{2}kd^2\)
3. \(mg(h-d)-\frac{1}{2}kd^2\)
4. \(mg(h-d)+\frac{1}{2}kd^2\)