Question

a body of mass 10 kg is moving with a velocity of 5 m/s attainsvof 35m/s in 5 second find the force applied on body to change its velocity



ans step by step​

Answer 1

Answer:

For a perfectly elastic collision, coefficient of restitution is e=1.

Let u and v be the velocities of 10 kg and 1 g masses after collision.

Thus, e=

rel. vel. of approach

rel. vel.of separation

=

5−0

v−u

=

5

v−u

But, as e=1, we have v−u=5.

Total momentum before collision is P=10×5=50 kg m/s

Total momentum after collision is P=10×(v−5)+0.001×v=10.001v−50

Thus, 10.001v−50=50⇒v=

10.001

100

≈10 m/s

Explanation:

please follow

Answer 2

Explanation:

GIVEN :-

• Initial velocity , u = 5 m/s.
• Final velocity , v = 35 m/s.

• t = 5 seconds.

• Mass of body , m = 10 kg.

TO FIND :-

• The force applied , F

SOLUTION :-

$\\ \underline{ \bigstar \: \textsf{Apply the 1st \: Equation of motion.}}$

$: \implies \displaystyle \sf \:v = u + at$

$: \implies \displaystyle \sf \: 35 \: ms ^{ - 1} = 5ms ^{ - 1} + a \times 5$

$: \implies \displaystyle \sf \:35 \: ms ^{ - 1} - 5 \: ms ^{ - 1} = 5s \times a$

$: \implies \displaystyle \sf \:30 \: ms ^{ - 1} = 5s \times a$

$: \implies \displaystyle \sf \:a = \dfrac{30 \: ms ^{ - 1} }{5}$

$: \implies \underline{ \boxed{\displaystyle \sf \bold{ \:a = 6 \: ms ^{ - 2} }}}$

__________________…

$\underline{ \bigstar \: \textsf{Apply the \: formula \: of \: force .}}$

$: \implies \displaystyle \sf \:Force = mass \times Acceleration$

$: \implies \displaystyle \sf \:Force =10 \: kg \times 6 \: ms^{ - 2}$

$: \implies \displaystyle \sf \:Force =60 \: kg.ms ^{ - 2}$

$: \implies \underline{ \boxed{ \displaystyle \sf \bold{ \:Force =60 \: N}}}$