Explanations & Calculations
- Say the velocities of A and B just after the collision are vavb .
- By conservation of linear momentum we get,
ma.(20)−mb.(30)va+vb→(Right is taken positive)=ma.(va)+mb.(vb)=−10
- By Newton's experimental law,
va−vbva−vb=−0.80[20−(−30)]=−40
- By solving the two simultaneous eq.s
vavb=−25ms−1=15ms−1
- They will move together attached as one after the collision in this case. Then by conservation of linear momentum,
ma(20)+mb(−30)V=(ma+mb)V=−5ms−1
Comments
Leave a comment