First slide

Motion of centre of mass

Question

Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass. Spring Constant of spring is K = 140 N/M and blocks are placed on a frictionless horizontal surface. An impulse gives a speed of 14 m/s to 10 kg block in the direction of the lighter block. Then, the maximum compression in the spring will be

Moderate

A

1 m
B

2 m
C

3 m
D

4 m

Solution

$\frac{1}{2} \frac{m_{1} m_{3}}{m_{1} + m_{2}} {(v_{1} - v_{2})}^{2} = \frac{1}{2} k x^{2}$
$\frac{1}{2} \times \frac{40}{14} (14)^{2} = \frac{1}{2} \times 140 x^{2}$
$4 = x^{2}$
X = 2m

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App