How Forces Affect Motion — CBSE Class 9 Science Important Questions

Answer: (b) $\text{kg m s}^{-1}$kg m s^-1.

Momentum $p=mv$p=mv, so its unit is (mass)$\times$×(velocity) $=\text{kg}\times\text{m s}^{-1}=\text{kg m s}^{-1}$=kg×m s^-1=kg m s^-1.

Answer: (b) Inertia.

Inertia is the natural tendency of an object to oppose a change in its state of motion; it is measured by the object's mass.

Answer: (a) $5\ \text{m s}^{-2}$5 m s^-2.

From $F=ma$F=ma, $a=\dfrac{F}{m}=\dfrac{10}{2}=5\ \text{m s}^{-2}$a=F/m=10/2=5 m s^-2.

Answer: (b) $\dfrac{mv}{M}$mv/M in the direction opposite to the bullet.

Initial momentum is zero. By conservation of momentum, $0=mv+MV$0=mv+MV, so $V=-\dfrac{mv}{M}$V=-mv/M; the negative sign shows the gun recoils opposite to the bullet.

Answer: (a) Both A and R are true and R is the correct explanation of A. Because of inertia of motion, the upper body continues to move forward when the bus stops, throwing the passengers forward.

Newton's second law: the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force, giving $F=ma$F=ma.

One newton is the force that produces an acceleration of $1\ \text{m s}^{-2}$1 m s^-2 in a body of mass $1\ \text{kg}$1 kg: $1\ \text{N}=1\ \text{kg m s}^{-2}$1 N=1 kg m s^-2.

Initial momentum $=mu=5\times4=20\ \text{kg m s}^{-1}$=mu=5×4=20 kg m s^-1; final momentum $=0$=0.

$F=\dfrac{mv-mu}{t}=\dfrac{0-20}{2}=-10\ \text{N}$F=mv-mu/t=0-20/2=-10 N.

The magnitude of the retarding force is $10\ \text{N}$10 N, acting opposite to the motion.

A fast ball has a large momentum $p=mv$p=mv that must be reduced to zero when caught.

By Newton's second law, $F=\dfrac{\text{change in momentum}}{\text{time}}$F=change in momentum/time. By drawing the hands backwards, the player increases the time $t$t over which the ball's momentum falls to zero.

Since the change in momentum is fixed, increasing $t$t decreases the force $F$F on the hands, so the catch hurts less and is safer.

Let a body of mass $m$m have initial velocity $u$u. A force $F$F acting for time $t$t changes its velocity to $v$v.

Initial momentum $=mu$=mu, final momentum $=mv$=mv. By the second law, force is proportional to the rate of change of momentum:
$F\propto\dfrac{mv-mu}{t}=\dfrac{m(v-u)}{t}.$F-mu/t=m(v-u)/t.

Since $\dfrac{v-u}{t}=a$v-u/t=a (acceleration),
$F\propto ma\implies F=kma.$F ma F=kma.

Units are chosen so that $k=1$k=1, giving
$F=ma.$F=ma.

Given $m=10\ \text{kg}$m=10 kg, $u=5\ \text{m s}^{-1}$u=5 m s^-1, $v=15\ \text{m s}^{-1}$v=15 m s^-1, $t=5\ \text{s}$t=5 s.

(i) Change in momentum $=mv-mu=m(v-u)=10\times(15-5)=100\ \text{kg m s}^{-1}$=mv-mu=m(v-u)=10×(15-5)=100 kg m s^-1.

(ii) $F=\dfrac{\text{change in momentum}}{t}=\dfrac{100}{5}=20\ \text{N}$F=change in momentum/t=100/5=20 N.

Law of conservation of momentum: in the absence of an external force, the total momentum of a system of bodies remains constant.

Derivation: Two bodies of masses $m_1$m_1 and $m_2$m_2 move along a line with velocities $u_1$u_1 and $u_2$u_2 ($u_1>u_2$u_1>u_2). They collide for time $t$t and then move with velocities $v_1$v_1 and $v_2$v_2.

Force on body 2 by body 1: $F_{12}=\dfrac{m_2v_2-m_2u_2}{t}$F_12=m_2v_2-m_2u_2/t.
Force on body 1 by body 2: $F_{21}=\dfrac{m_1v_1-m_1u_1}{t}$F_21=m_1v_1-m_1u_1/t.

By Newton's third law these are equal and opposite: $F_{12}=-F_{21}$F_12=-F_21.
$\dfrac{m_2v_2-m_2u_2}{t}=-\dfrac{m_1v_1-m_1u_1}{t}.$m_2v_2-m_2u_2/t=-m_1v_1-m_1u_1/t.

Cancelling $t$t and rearranging:
$m_1u_1+m_2u_2=m_1v_1+m_2v_2.$m_1u_1+m_2u_2=m_1v_1+m_2v_2.

Thus total momentum before collision equals total momentum after collision.

(a) Newton's third law: to every action there is an equal and opposite reaction, and the two act on different bodies. Example: when we walk, our foot pushes the ground backward (action) and the ground pushes us forward (reaction).

(b) The boy and ball are initially at rest, so total initial momentum $=0$=0. Let the boy's recoil velocity be $V$V; take the ball's direction as positive.

By conservation of momentum:
$0=m_{\text{ball}}v_{\text{ball}}+m_{\text{boy}}V$0=m_ballv_ball+m_boyV
$0=(2)(5)+(60)V$0=(2)(5)+(60)V
$V=-\dfrac{10}{60}=-0.167\ \text{m s}^{-1}.$V=-10/60=-0.167 m s^-1.

The boy recoils at about $0.17\ \text{m s}^{-1}$0.17 m s^-1 in the direction opposite to the ball.

(i) The inertia of motion of the passenger's body makes it tend to continue moving forward.

(ii) The momentum $p=mv$p=mv of the passenger must be reduced to zero.

(iii) The change in momentum in a crash is fixed. Since $F=\dfrac{\text{change in momentum}}{\text{time}}$F=change in momentum/time, increasing the stopping time $t$t (as seat belts and air bags do) decreases the force $F$F experienced by the passenger, reducing injury.

(iv) Newton's first law of motion (the law of inertia) explains why the body continues to move forward when the car suddenly stops.