SB TEST PRO HUB

JEE Main Past Year Questions With Solutions on Kinematics 1D

Q1: A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 Kg is fired vertically upward, with a velocity 100 m/s, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is (g = 10 m/s)

1. 10 m
2. 30 m
3. 20 m
4. 40 m

Solution

Suppose both collide at the point P after time t. Time taken for the particles to collide,

t = d/vrel = 100/100 = 1s

Speed of wood just before collision =gt = 10m/s

Speed of bullet just before collision

v -gt = 100 -10 = 90 m/s

Before

0.03 kg ↓ 10 m/s

0.02 kg ↑ 90 m/s

After

↑ v

0.05 kg

Now, the conservation of linear momentum just before and after the collision

-(0.03)(10) + (0.02)(90) = (0.05)v ⇒ v = 30 m/s

The maximum height reached by the body a = v2/2g

= (30)2/2(10)

= 45 m

(100 -h) = ½ gt2 = ½ x 10 x1 ⇒h = 95 m

Height above tower = 40 m

Answer: (d) 40 m

Q2: A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of 30 km/hr. The ratio of times taken by the passenger train to completely cross the freight train when: (i) they are moving in the same direction, and (ii) in the opposite direction, is

(a) 25/11

(b) 3/2

(c) 5/2

(d) 11/5

Solution:

The total distance to be travelled by train is 60 + 120 = 180 m.

When the trains are moving in the same direction, the relative velocity is v1 – v2 = 80 – 30 = 50 km hr–1.

So time taken to cross each other, t1 = 180/(50 x 103/3600) = [(18 x 18)/25] s

When the trains are moving in opposite direction, relative velocity is |v1 – (–v2 )| = 80 + 30 = 110 km hr–1

So time taken to cross each other

t2= 180/(110 x 103/3600) = [(18 x 36)/110] s

t1/t2= [(18 x 18)/25] / [(18 x 36)/110] = 11/5

Answers: (d) 11/5

Q3: A particle has an initial velocity

\(\begin{array}{l}3{\hat{i}}+4{\hat{j}}\end{array} \)

and an acceleration of

\(\begin{array}{l}0.4{\hat{i}}+0.3{\hat{j}}\end{array} \)

. Its speed after 10s is

1. 7 units
2. 8.5 units
3. 10 units
4. 7√2 units

Solution:

v = u + at

v = 3i + 4j + (0.4 i + 0.3 j) x 10

= 3i + 4j + 4i + 3i

v = 7i + 7j

\(\begin{array}{l}\left | \vec{v} \right |= \sqrt{7^{2}+7^{2}}\end{array} \)

Answer: (d) 7√2 units

Q4: An automobile travelling at 40 km/h can be stopped at a distance of 40 m by applying brakes. If the same automobile is travelling at 80 km/h, the minimum stopping distance, in metres, is (assume no skidding)

(a) 100 m

(b) 75 m

(c) 160 m

(d) 150 m

Solution :

Using v2 = u2 – 2as

0 = u2 – 2as

S = u2 /2a

S1/S2 = u12/u22

S2 = (u12/u22)S1 = (2)2(40) = 160 m

Answer: (c)160 m