Question

27. Figure shows plot of force magnitude vs time
during the collision of a 50 g ball with wall. The
initial velocity of the ball is 30 m/s perpendicular to
the wall, the ball rebounds directly back with same
speed perpendicular to the wall. The maximum
magnitude of the force on the ball from the wall
during the collision is
F(N)
max
2
4
6
(ms)
(1) 1000 N
12V 750 N
(3) 1500 N
(4) 2000 N​

Answer 1

I hope that this will help you alot...!!!

Answer:

(3)..1500N

Explanation:

1) We have, a 50 g ball moving with initial velocity 30 m/s perpendicular to wall.

We know,

By Newton's Second Law,

Rate of change of momentum is Force.

That is,

F=\frac{dp}{dt}F=

dt

dp

2) Now,

F_max acts for (4-2) = 2ms

Since, direction of velocity changes when ball bounces back with same speed.

\begin{gathered}=>\Delta p = mv-(-mv)=2mv\\ \\=2*50*10^{-3}*30=3 kgm/s\end{gathered}

=>Δp=mv−(−mv)=2mv

=2∗50∗10

−3

∗30=3kgm/s

3) \Delta t=2msΔt=2ms

That is,

F_{mag}=\frac{\Delta p}{\Delta t} =\frac{3}{2*10^{-3}}=1500NF

mag

=

Δt

Δp

=

2∗10

−3

3

=1500N

Hence, Value of F_max is 1500 N.