If two bodies having masses m1 and m2 (m1 > m2) have equal kinetic energies, the momentum of body having mass m1 is _________ the momentum of body having mass m2.
A. Equal to
B. Less than
C. Greater than
D. None of these
Answer: Option C
This Question Belongs to Mechanical Engineering >> Engineering Mechanics
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The resultant of two equal forces P making an angle $$\theta ,$$ is given by
A. $$2{\text{P}}\sin \frac{\theta }{2}$$
B. $$2{\text{P}}\cos \frac{\theta }{2}$$
C. $$2{\text{P}}\tan \frac{\theta }{2}$$
D. $$2{\text{P}}\cot \frac{\theta }{2}$$
A. Equal to
B. Less than
C. Greater than
D. None of these
If a number of forces are acting at a point, their resultant is given by
A. $${\left( {\sum {\text{V}} } \right)^2} + {\left( {\sum {\text{H}} } \right)^2}$$
B. $$\sqrt {{{\left( {\sum {\text{V}} } \right)}^2} + {{\left( {\sum {\text{H}} } \right)}^2}} $$
C. $${\left( {\sum {\text{V}} } \right)^2} + {\left( {\sum {\text{H}} } \right)^2} + 2\left( {\sum {\text{V}} } \right)\left( {\sum {\text{H}} } \right)$$
D. $$\sqrt {{{\left( {\sum {\text{V}} } \right)}^2} + {{\left( {\sum {\text{H}} } \right)}^2} + 2\left( {\sum {\text{V}} } \right)\left( {\sum {\text{H}} } \right)} $$
A. $${\text{a}} = \frac{\alpha }{{\text{r}}}$$
B. $${\text{a}} = \alpha {\text{r}}$$
C. $${\text{a}} = \frac{{\text{r}}}{\alpha }$$
D. None of these
m1v1²=m2v2²
m1/m2=(V2/V1)²
m1/M2>1 so V2/V1 >1
Momentum ratio =m1v1/m2v2
Substitute m1/M2=(V2/V1)² in momentum ratio
Momentum ratio =(V2/V1)² ×(V1/V2)=V2/V1 which is greater than one
m1>m2
KE 1 = KE 2
m1v1² = m2v2²
(v2/v1) = √(m1/m2)
Momentum(p) = mv
KE = 0.5 mv² = 0.5 pv
p1 = ( v2/v1) p2
p1 √m2 =p2 √ m1
As m1>m2
p1
how explain